Electrostatics

Introduction & Properties of Charge

Electrostatics deals with the study of forces, fields, and potentials arising from static (stationary) electric charges.

Electric Charge

Electric charge is the fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. It is a scalar quantity.

There are two kinds of charges: positive (+) and negative (-).
By convention, the charge on a glass rod rubbed with silk is positive, and the charge on an electron is negative.
Like charges repel, and unlike charges attract each other.

Conductors and Insulators

Conductors: Materials that allow electric charge (usually electrons) to move freely through them. Examples: metals, human body, Earth.
Insulators: Materials that offer high resistance to the movement of charge. Examples: glass, plastic, rubber, wood.
Charging by Induction: A charged object brought near a neutral conductor will cause the charges in the conductor to separate. This allows for charging the conductor without direct contact.

Basic Properties of Electric Charge

1. Additivity of Charges

The total charge of a system is the algebraic sum of all the individual charges in the system. If a system contains $n$ charges $q_{1}, q_{2}, . . ., q_{n}$ , the total charge $Q$ is:

Q = q_{1} + q_{2} + q_{3} + . . . + q_{n}

2. Conservation of Charge

The total charge of an isolated system remains constant. Charge can neither be created nor destroyed; it can only be transferred from one body to another.

3. Quantization of Charge

All free charges are integral multiples of a basic unit of charge, denoted by $e$ . The charge $q$ on any body is given by:

q = n e

where $n$ is an integer ( $n = 0, \pm 1, \pm 2, . . .$ ). The basic unit of charge is the magnitude of the charge on a single electron or proton.

e = 1.602 \times 10^{- 19} C

The SI unit of charge is the Coulomb (C).

Coulomb's Law

Coulomb's Law gives the magnitude of the electrostatic force between two stationary point charges.

The force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

F = k \frac{| q_{1} q_{2} |}{r^{2}}

where:

$F$ is the magnitude of the electrostatic force.
$q_{1}$ and $q_{2}$ are the magnitudes of the point charges.
$r$ is the distance between the charges.
$k$ is the electrostatic constant, also written as $\frac{1}{4 π ϵ_{0}}$ .

In SI units, the value of $k$ is approximately $9 \times 10^{9} {N m}^{2} C^{- 2}$ .

$ϵ_{0}$ is the permittivity of free space, with a value of $8.854 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ .

Coulomb's Law in Vector Form

The force is a vector quantity. The force on charge $q_{2}$ due to $q_{1}$ , denoted by ${\vec{F}}_{21}$ , is given by:

{\vec{F}}_{21} = k \frac{q_{1} q_{2}}{r_{21}^{2}} {\hat{r}}_{21} = \frac{1}{4 π ϵ_{0}} \frac{q_{1} q_{2}}{| {\vec{r}}_{2} - {\vec{r}}_{1} |^{3}} ({\vec{r}}_{2} - {\vec{r}}_{1})

where ${\hat{r}}_{21}$ is the unit vector pointing from $q_{1}$ to $q_{2}$ .

If $q_{1} q_{2} > 0$ (like charges), the force is repulsive (in the direction of ${\hat{r}}_{21}$ ).
If $q_{1} q_{2} < 0$ (unlike charges), the force is attractive (opposite to the direction of ${\hat{r}}_{21}$ ).

By Newton's third law, the force on $q_{1}$ due to $q_{2}$ is equal and opposite: ${\vec{F}}_{12} = - {\vec{F}}_{21}$ .

Forces between Multiple Charges: Superposition Principle

The principle of superposition states that the total force on any given charge due to a number of other charges is the vector sum of all the individual forces exerted on it by the other charges.

The force between any two charges is independent of the presence of other charges.

For a system of charges $q_{1}, q_{2}, . . ., q_{n}$ , the total force ${\vec{F}}_{1}$ on charge $q_{1}$ is:

{\vec{F}}_{1} = {\vec{F}}_{12} + {\vec{F}}_{13} + . . . + {\vec{F}}_{1 n} = \sum_{i = 2}^{n} {\vec{F}}_{1 i}

{\vec{F}}_{1} = \frac{1}{4 π ϵ_{0}} \sum_{i = 2}^{n} \frac{q_{1} q_{i}}{r_{1 i}^{2}} {\hat{r}}_{1 i}

Electric Field

The electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.

The electric field $\vec{E}$ at a point is defined as the electrostatic force $\vec{F}$ experienced by a small positive test charge $q_{0}$ placed at that point, divided by the magnitude of the test charge.

\vec{E} = lim_{q_{0} \to 0} \frac{\vec{F}}{q_{0}}

The SI unit for electric field is Newtons per Coulomb (N/C).

Electric Field due to a Point Charge

The electric field $\vec{E}$ created by a source charge $q$ at a distance $r$ from it is:

\vec{E} = \frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}} \hat{r}

The direction of $\vec{E}$ is radially outward from a positive charge and radially inward toward a negative charge.

Electric Field due to a System of Charges

Similar to force, the electric field at a point due to a system of charges is the vector sum of the electric fields at that point due to each individual charge.

{\vec{E}}_{t o t a l} = {\vec{E}}_{1} + {\vec{E}}_{2} + . . . + {\vec{E}}_{n} = \sum_{i = 1}^{n} {\vec{E}}_{i}

{\vec{E}}_{t o t a l} = \frac{1}{4 π ϵ_{0}} \sum_{i = 1}^{n} \frac{q_{i}}{r_{i}^{2}} {\hat{r}}_{i}

Electric Field Lines

An electric field line is a curve drawn in such a way that the tangent to it at each point is in the direction of the net electric field at that point. They are a visual way to map electric fields.

Properties of Electric Field Lines:

Field lines start from positive charges and end at negative charges. If there is a single charge, they may start or end at infinity.
In a charge-free region, electric field lines can be taken to be continuous curves without any breaks.
Two field lines can never cross each other. If they did, the field at the point of intersection would have two directions, which is impossible.
Electrostatic field lines do not form any closed loops. This is a consequence of the conservative nature of the electrostatic field.
The density of the field lines (number of lines per unit area perpendicular to the lines) is proportional to the magnitude of the electric field strength. Closer lines mean a stronger field.

Electric Dipole

An electric dipole is a pair of equal and opposite point charges, $+ q$ and $- q$ , separated by a small distance $2 a$ .

Dipole Moment

The electric dipole moment, $\vec{p}$ , is a vector quantity used to measure the strength of an electric dipole. Its magnitude is the product of the charge and the separation distance, and its direction is from the negative charge to the positive charge.

| \vec{p} | = q (2 a)

\vec{p} = q (2 \vec{a})

Electric Field of a Dipole

1. Field on the Axial Line

The electric field at a point P on the axis of the dipole at a distance $r$ from the center is:

{\vec{E}}_{a x i a l} = \frac{1}{4 π ϵ_{0}} \frac{2 p r}{(r^{2} - a^{2})^{2}} \hat{p}

For a short dipole where $r ≫ a$ , this simplifies to:

{\vec{E}}_{a x i a l} \approx \frac{1}{4 π ϵ_{0}} \frac{2 \vec{p}}{r^{3}}

2. Field on the Equatorial Line

The electric field at a point P on the equatorial plane at a distance $r$ from the center is:

{\vec{E}}_{e q u a t o r i a l} = - \frac{1}{4 π ϵ_{0}} \frac{\vec{p}}{(r^{2} + a^{2})^{3 / 2}}

For a short dipole where $r ≫ a$ , this simplifies to:

{\vec{E}}_{e q u a t o r i a l} \approx - \frac{1}{4 π ϵ_{0}} \frac{\vec{p}}{r^{3}}

Note that the dipole field falls off as $1 / r^{3}$ , much faster than the $1 / r^{2}$ field of a single point charge.

Torque on a Dipole in a Uniform Electric Field

When a dipole is placed in a uniform external electric field $\vec{E}$ , it experiences a torque $\vec{τ}$ that tends to align it with the field. The net force on the dipole is zero.

\vec{τ} = \vec{p} \times \vec{E}

The magnitude of the torque is $τ = p E \sin θ$ , where $θ$ is the angle between $\vec{p}$ and $\vec{E}$ .

Torque is maximum ( $τ = p E$ ) when $θ = 90^{\circ}$ (dipole is perpendicular to the field).
Torque is zero ( $τ = 0$ ) when $θ = 0^{\circ}$ (stable equilibrium) or $θ = 180^{\circ}$ (unstable equilibrium).

Continuous Charge Distribution

For macroscopic charge distributions, it's often convenient to treat the charge as continuous.

Linear Charge Density ( $λ$ ): Charge per unit length. Unit: C/m. $d q = λ d l$ .
Surface Charge Density ( $σ$ ): Charge per unit area. Unit: C/m². $d q = σ d S$ .
Volume Charge Density ( $ρ$ ): Charge per unit volume. Unit: C/m³. $d q = ρ d V$ .

The electric field from a continuous distribution is found by integrating the contributions from infinitesimal charge elements $d q$ :

\vec{E} = \frac{1}{4 π ϵ_{0}} \int \frac{d q}{r^{2}} \hat{r}

Electric Flux

Electric flux ( $Φ_{E}$ ) is a measure of the "flow" of the electric field through a given area. For a uniform electric field $\vec{E}$ passing through a small planar area element $Δ \vec{S}$ , the flux is:

Δ Φ_{E} = \vec{E} \cdot Δ \vec{S} = E Δ S \cos θ

where $θ$ is the angle between the electric field vector and the area vector (which is normal to the surface). The SI unit of electric flux is N m²/C.

For a general surface, the total flux is found by integrating over the entire surface:

Φ_{E} = \oint \vec{E} \cdot d \vec{S}

The circle on the integral sign indicates integration over a closed surface. For a closed surface, the area vector $d \vec{S}$ is conventionally taken to point outward.

Gauss's Law & Applications

Gauss's Law

Gauss's Law states that the total electric flux through any closed surface (called a Gaussian surface) is equal to $1 / ϵ_{0}$ times the net electric charge $q_{e n c}$ enclosed by the surface.

Φ_{E} = \oint \vec{E} \cdot d \vec{S} = \frac{q_{e n c}}{ϵ_{0}}

Gauss's law is extremely useful for calculating the electric field for charge distributions with high symmetry (spherical, cylindrical, or planar).

Applications of Gauss's Law

1. Field due to an Infinitely Long Straight Uniformly Charged Wire

Consider a wire with uniform linear charge density $λ$ . By symmetry, the electric field must be radial and perpendicular to the wire.

We choose a cylindrical Gaussian surface of radius $r$ and length $l$ , coaxial with the wire.

Flux through the top and bottom circular caps is zero as $\vec{E}$ is perpendicular to $d \vec{S}$ .
Flux through the curved surface: $\vec{E}$ is parallel to $d \vec{S}$ , so $Φ = E \cdot (2 π r l)$ .

The charge enclosed is $q_{e n c} = λ l$ . Applying Gauss's Law:

E (2 π r l) = \frac{λ l}{ϵ_{0}} ⟹ E = \frac{λ}{2 π ϵ_{0} r}

2. Field due to a Uniformly Charged Infinite Plane Sheet

Consider an infinite plane sheet with uniform surface charge density $σ$ . By symmetry, the electric field must be perpendicular to the sheet, pointing away from it (if $σ > 0$ ).

We choose a cylindrical or rectangular "pillbox" Gaussian surface that pierces the sheet, with its flat ends of area $A$ parallel to the sheet.

Flux through the curved sides is zero as $\vec{E}$ is perpendicular to $d \vec{S}$ .
Flux through the two flat ends adds up: $Φ = E A + E A = 2 E A$ .

The charge enclosed is $q_{e n c} = σ A$ . Applying Gauss's Law:

2 E A = \frac{σ A}{ϵ_{0}} ⟹ E = \frac{σ}{2 ϵ_{0}}

Interestingly, the field is uniform and does not depend on the distance from the sheet.

3. Field due to a Uniformly Charged Thin Spherical Shell

Consider a thin spherical shell of radius $R$ with total charge $q$ uniformly distributed on its surface ( $σ = q / 4 π R^{2}$ ).

Case 1: Outside the shell ( $r > R$ )
We choose a spherical Gaussian surface of radius $r > R$ . The enclosed charge is $q$ .
$E (4 π r^{2}) = \frac{q}{ϵ_{0}} ⟹ E = \frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}}$
The field outside is the same as if all the charge were a point charge at the center.
Case 2: Inside the shell ( $r < R$ )
We choose a spherical Gaussian surface of radius $r < R$ . The enclosed charge is zero.
$E (4 π r^{2}) = \frac{0}{ϵ_{0}} ⟹ E = 0$
The electric field is zero everywhere inside a uniformly charged spherical shell.

NCERT Examples Solved

Example 1.1

Question

If $10^{9}$ electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?

Solution

The charge on a single electron is $e = 1.6 \times 10^{- 19}$ C.

Number of electrons moving per second, $n = 10^{9}$ electrons/s.

The total charge moving out of the body in one second is:

Δ q = n \times e = 10^{9} \times (1.6 \times 10^{- 19} C) = 1.6 \times 10^{- 10} C/s

We need to find the time ( $t$ ) required to accumulate a total charge ( $Q$ ) of 1 C.

t = \frac{Q}{Δ q} = \frac{1 C}{1.6 \times 10^{- 10} C/s} = 6.25 \times 10^{9} s

To convert this time into years:

t = \frac{6.25 \times 10^{9} s}{365 days/year \times 24 hours/day \times 3600 s/hour} \approx 198 years

Therefore, it would take approximately 198 years to accumulate a charge of 1 C.

Example 1.2

Question

How much positive and negative charge is there in a cup of water?

Solution

Let's assume the mass of water in one cup is $m = 250$ g.

The molecular mass of water (H₂O) is $M = 1 \times 2 + 16 = 18$ g/mol.

The number of moles of water in the cup is:

moles = \frac{m}{M} = \frac{250 g}{18 g/mol}

The number of molecules of water ( $N$ ) in the cup, using Avogadro's number ( $N_{A} = 6.022 \times 10^{23}$ molecules/mol), is:

N = \frac{250}{18} \times 6.022 \times 10^{23} molecules

Each molecule of water (H₂O) contains 2 hydrogen atoms and 1 oxygen atom.

Number of electrons = $2 (1) + 8 = 10$ electrons.
Number of protons = $2 (1) + 8 = 10$ protons.

The total positive charge (from protons) or total negative charge (from electrons) is given by $Q = (Total number of particles) \times e$ .

Q = N \times 10 \times e = (\frac{250}{18} \times 6.022 \times 10^{23}) \times 10 \times (1.6 \times 10^{- 19} C)

Q \approx 1.34 \times 10^{7} C

Thus, a cup of water contains approximately 1.34 × 10⁷ C of positive charge and an equal amount of negative charge.

Example 1.3

Question

Coulomb's law for electrostatic force and Newton's law for gravitational force both have inverse-square dependence on distance.
(a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons.
(b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Å apart.

Solution (a)

Electrostatic Force: $F_{e} = k \frac{| q_{1} q_{2} |}{r^{2}}$

Gravitational Force: $F_{g} = G \frac{m_{1} m_{2}}{r^{2}}$

Constants: $e = 1.6 \times 10^{- 19}$ C, $m_{e} = 9.11 \times 10^{- 31}$ kg, $m_{p} = 1.67 \times 10^{- 27}$ kg, $k = 9 \times 10^{9}$ N m²/C², $G = 6.67 \times 10^{- 11}$ N m²/kg².

(i) For an electron and a proton:

\frac{F_{e}}{F_{g}} = \frac{k e^{2} / r^{2}}{G m_{p} m_{e} / r^{2}} = \frac{k e^{2}}{G m_{p} m_{e}}

\frac{F_{e}}{F_{g}} = \frac{(9 \times 10^{9}) (1.6 \times 10^{- 19})^{2}}{(6.67 \times 10^{- 11}) (1.67 \times 10^{- 27}) (9.11 \times 10^{- 31})} \approx 2.27 \times 10^{39}

(ii) For two protons:

\frac{F_{e}}{F_{g}} = \frac{k e^{2} / r^{2}}{G m_{p} m_{p} / r^{2}} = \frac{k e^{2}}{G m_{p}^{2}}

\frac{F_{e}}{F_{g}} = \frac{(9 \times 10^{9}) (1.6 \times 10^{- 19})^{2}}{(6.67 \times 10^{- 11}) (1.67 \times 10^{- 27})^{2}} \approx 1.24 \times 10^{36}

These ratios show that the electrostatic force is enormously stronger than the gravitational force.

Solution (b)

The distance is $r = 1$ Å = $10^{- 10}$ m.

The magnitude of the electric force between the electron and proton is:

F_{e} = k \frac{e^{2}}{r^{2}} = (9 \times 10^{9}) \frac{(1.6 \times 10^{- 19})^{2}}{(10^{- 10})^{2}} \approx 2.3 \times 10^{- 8} N

Using Newton's second law, $F = m a$ , we find the accelerations.

Acceleration of the electron ( $a_{e}$ ):

a_{e} = \frac{F_{e}}{m_{e}} = \frac{2.3 \times 10^{- 8} N}{9.11 \times 10^{- 31} kg} \approx 2.5 \times 10^{22} {m/s}^{2}

Acceleration of the proton ( $a_{p}$ ):

a_{p} = \frac{F_{e}}{m_{p}} = \frac{2.3 \times 10^{- 8} N}{1.67 \times 10^{- 27} kg} \approx 1.4 \times 10^{19} {m/s}^{2}

These are extremely large accelerations, highlighting the dominance of the electric force at the atomic scale.

Example 1.4

Question

A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm. The resulting repulsion of A is noted. Spheres A and B are touched by uncharged spheres C and D respectively. C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres. What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres.

Solution

Let the initial charge on sphere A be $q_{A}$ and on sphere B be $q_{B}$ . The initial distance is $r_{1} = 10$ cm = $0.1$ m. The initial repulsive force is:

F_{1} = k \frac{q_{A} q_{B}}{r_{1}^{2}}

When sphere A is touched by an identical uncharged sphere C, the charge $q_{A}$ is shared equally. The new charge on A is $q_{A}^{'} = q_{A} / 2$ .

Similarly, when sphere B is touched by an identical uncharged sphere D, its new charge becomes $q_{B}^{'} = q_{B} / 2$ .

The new distance between A and B is $r_{2} = 5$ cm = $0.05$ m.

The new repulsive force $F_{2}$ is:

F_{2} = k \frac{q_{A}^{'} q_{B}^{'}}{r_{2}^{2}} = k \frac{(q_{A} / 2) (q_{B} / 2)}{(0.05)^{2}} = k \frac{q_{A} q_{B}}{4 \times (0.05)^{2}}

Let's compare this with the original force. We have $r_{2} = r_{1} / 2$ .

F_{2} = k \frac{(q_{A} / 2) (q_{B} / 2)}{(r_{1} / 2)^{2}} = k \frac{q_{A} q_{B} / 4}{r_{1}^{2} / 4} = k \frac{q_{A} q_{B}}{r_{1}^{2}} = F_{1}

The expected repulsion force on A remains unchanged. It is equal to the initial force $F_{1}$ .

Example 1.5

Question

Consider three charges $q_{1}, q_{2}, q_{3}$ each equal to $q$ at the vertices of an equilateral triangle of side $l$ . What is the force on a charge $Q$ (with the same sign as $q$ ) placed at the centroid of the triangle?

Solution

Let the vertices be A, B, C with charges $q_{1}, q_{2}, q_{3}$ respectively, and the centroid be O with charge $Q$ . In an equilateral triangle, the centroid O is equidistant from all three vertices.

The distance $r$ from any vertex to the centroid is $r = \frac{l}{\sqrt{3}}$ .

The magnitude of the force exerted by each vertex charge on the charge $Q$ at the centroid is the same:

| {\vec{F}}_{O A} | = | {\vec{F}}_{O B} | = | {\vec{F}}_{O C} | = F = k \frac{| q Q |}{r^{2}} = k \frac{| q Q |}{(l / \sqrt{3})^{2}} = \frac{3 k | q Q |}{l^{2}}

The forces are directed away from the respective vertices (since $q$ and $Q$ have the same sign). Let's denote the forces as ${\vec{F}}_{A}, {\vec{F}}_{B}, {\vec{F}}_{C}$ . These three force vectors are equal in magnitude and are directed 120° apart from each other.

By symmetry, the vector sum of these three forces is zero. We can also show this by resolving components. Let ${\vec{F}}_{A}$ be along the negative y-axis. Then ${\vec{F}}_{B}$ and ${\vec{F}}_{C}$ will be at 120° and 240° to ${\vec{F}}_{A}$ . The sum of the forces is:

{\vec{F}}_{n e t} = {\vec{F}}_{A} + {\vec{F}}_{B} + {\vec{F}}_{C}

The y-components: $- F + F \cos (120^{\circ}) + F \cos (120^{\circ}) = - F + F (- 1 / 2) + F (- 1 / 2) = - F + F = 0$ .
The x-components: $0 + F \sin (120^{\circ}) + F \sin (- 120^{\circ}) = F (\sqrt{3} / 2) - F (\sqrt{3} / 2) = 0$ .

Thus, the net force on the charge $Q$ at the centroid is zero.

Example 1.6

Question

Consider the charges q, q, and –q placed at the vertices of an equilateral triangle. What is the force on each charge?

Solution

Let the charges be at vertices A ( $+ q$ ), B ( $+ q$ ), and C ( $- q$ ) of an equilateral triangle with side length $l$ . The magnitude of the force between any two charges is $F = k \frac{q^{2}}{l^{2}}$ .

Force on charge at A ( $+ q$ ):

Force from B ( ${\vec{F}}_{A B}$ ): Repulsive, directed along BA. Magnitude is $F$ .
Force from C ( ${\vec{F}}_{A C}$ ): Attractive, directed along AC. Magnitude is $F$ .
The angle between ${\vec{F}}_{A B}$ and ${\vec{F}}_{A C}$ is 120°. The resultant force $| {\vec{F}}_{A} |$ is:

| {\vec{F}}_{A} | = \sqrt{F^{2} + F^{2} + 2 F^{2} \cos (120^{\circ})} = \sqrt{2 F^{2} + 2 F^{2} (- 1 / 2)} = \sqrt{F^{2}} = F

The direction of ${\vec{F}}_{A}$ is parallel to the side BC.

Force on charge at B ( $+ q$ ):

By symmetry with vertex A, the situation is identical. The force from A is repulsive (along AB), and the force from C is attractive (along BC). The angle is again 120°.
The resultant force $| {\vec{F}}_{B} | = F$ . The direction is parallel to the side AC.

Force on charge at C ( $- q$ ):

Force from A ( ${\vec{F}}_{C A}$ ): Attractive, directed along CA. Magnitude is $F$ .
Force from B ( ${\vec{F}}_{C B}$ ): Attractive, directed along CB. Magnitude is $F$ .
The angle between these two forces is 60°. The resultant force $| {\vec{F}}_{C} |$ is:

| {\vec{F}}_{C} | = \sqrt{F^{2} + F^{2} + 2 F^{2} \cos (60^{\circ})} = \sqrt{2 F^{2} + 2 F^{2} (1 / 2)} = \sqrt{3 F^{2}} = F \sqrt{3}

The direction of ${\vec{F}}_{C}$ bisects the angle at C and points inwards.

NCERT Exercises Solved

Exercise 1.1

Question

What is the force between two small charged spheres having charges of $2 \times 10^{- 7}$ C and $3 \times 10^{- 7}$ C placed 30 cm apart in air?

Solution

Given charges $q_{1} = 2 \times 10^{- 7}$ C, $q_{2} = 3 \times 10^{- 7}$ C.

Distance $r = 30$ cm = $0.3$ m.

Using Coulomb's Law:

F = k \frac{q_{1} q_{2}}{r^{2}}

Substituting the values, with $k = 9 \times 10^{9}$ N m²/C²:

F = (9 \times 10^{9}) \frac{(2 \times 10^{- 7}) (3 \times 10^{- 7})}{(0.3)^{2}} = (9 \times 10^{9}) \frac{6 \times 10^{- 14}}{0.09}

F = \frac{54 \times 10^{- 5}}{0.09} = 600 \times 10^{- 5} = 6 \times 10^{- 3} N

Since both charges are positive, the force is repulsive.

Exercise 1.2

Question

The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Solution

(a) Given $q_{1} = 0.4 μ C = 4 \times 10^{- 7}$ C, $q_{2} = - 0.8 μ C = - 8 \times 10^{- 7}$ C, and $F = 0.2$ N. The force is attractive.

From Coulomb's Law, $F = k \frac{| q_{1} q_{2} |}{r^{2}}$ . We can solve for $r$ :

r = \sqrt{k \frac{| q_{1} q_{2} |}{F}} = \sqrt{(9 \times 10^{9}) \frac{(4 \times 10^{- 7}) (8 \times 10^{- 7})}{0.2}}

r = \sqrt{\frac{28.8 \times 10^{- 4}}{0.2}} = \sqrt{144 \times 10^{- 4}} = 12 \times 10^{- 2} m = 12 cm

The distance between the spheres is 12 cm.

(b) According to Newton's Third Law, the force exerted by the first sphere on the second is equal in magnitude and opposite in direction to the force exerted by the second on the first. Therefore, the force on the second sphere is 0.2 N (attractive).

Exercise 1.3

Question

Check that the ratio $k e^{2} / (G m_{e} m_{p})$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Solution

Dimensionless Check:
The ratio is of the electrostatic force $F_{e} = k \frac{e^{2}}{r^{2}}$ to the gravitational force $F_{g} = G \frac{m_{e} m_{p}}{r^{2}}$ between an electron and a proton.
Since the ratio is a ratio of two forces, it is dimensionless.
Alternatively, let's check units:

\frac{[k] [e^{2}]}{[G] [m_{e}] [m_{p}]} = \frac{({N m}^{2} / C^{2}) (C^{2})}{({N m}^{2} / {kg}^{2}) (kg \cdot kg)} = \frac{{N m}^{2}}{{N m}^{2}} = 1

The ratio is indeed dimensionless.

Value of the Ratio:
This was calculated in Example 1.3.

\frac{k e^{2}}{G m_{e} m_{p}} = \frac{(9 \times 10^{9}) (1.6 \times 10^{- 19})^{2}}{(6.67 \times 10^{- 11}) (9.11 \times 10^{- 31}) (1.67 \times 10^{- 27})} \approx 2.27 \times 10^{39}

Significance:
This ratio represents the relative strength of the electrostatic force to the gravitational force between an electron and a proton. The enormously large value signifies that the electrostatic force is vastly stronger than the gravitational force at the atomic and subatomic levels.

Exercise 1.4

Question

(a) Explain the meaning of the statement 'electric charge of a body is quantised'.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Solution

(a) The statement 'electric charge of a body is quantised' means that any observable amount of electric charge ( $q$ ) is always an integer multiple of a fundamental unit of charge, denoted by $e$ (the magnitude of the charge of an electron). This can be expressed as $q = n e$ , where $n$ is an integer ( $0, \pm 1, \pm 2, . . .$ ). Charge is not continuous but exists in discrete packets.

(b) The basic unit of charge, $e = 1.6 \times 10^{- 19}$ C, is extremely small. In macroscopic situations, we deal with charges that are very large compared to $e$ (e.g., a charge of 1 µC contains about $6.25 \times 10^{12}$ elementary charges). Because the number of individual charges is so vast, the "steps" between allowed charge values are infinitesimally small compared to the total charge. This makes the charge appear to be a continuous quantity, and its discrete, or "grainy," nature can be safely ignored.

Exercise 1.5

Question

When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Solution

The law of conservation of charge states that the total charge of an isolated system remains constant. Charging by rubbing (frictional electricity) is a process of charge transfer, not charge creation.

When a glass rod is rubbed with silk, electrons are transferred from the glass to the silk.

The glass rod loses a certain number of electrons, so it acquires a net positive charge.
The silk cloth gains the exact same number of electrons, so it acquires a net negative charge of equal magnitude.

Before rubbing, both the rod and the cloth were electrically neutral (total charge = 0). After rubbing, the total charge of the system (rod + cloth) is $(+ q) + (- q) = 0$ . Since the total charge of the isolated system remains zero before and after the process, this observation is perfectly consistent with the law of conservation of charge.

Exercise 1.6

Question

Four point charges $q_{A} = 2 μ$ C, $q_{B} = - 5 μ$ C, $q_{C} = 2 μ$ C, and $q_{D} = - 5 μ$ C are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of $1 μ$ C placed at the centre of the square?

Solution

Let the square be ABCD with side length $s = 10$ cm = $0.1$ m. Let the charge at the center O be $q_{O} = 1 μ$ C.

The distance of the center O from each corner is the same. Due to the symmetry of the arrangement, we can analyze the forces in pairs.

Force due to $q_{A}$ and $q_{C}$ :
The charges $q_{A} = 2 μ$ C and $q_{C} = 2 μ$ C are equal and are at opposite corners. The force on $q_{O}$ due to $q_{A}$ ( ${\vec{F}}_{O A}$ ) is repulsive and directed along the diagonal from A to C. The force on $q_{O}$ due to $q_{C}$ ( ${\vec{F}}_{O C}$ ) is also repulsive and directed along the diagonal from C to A. Since the charges and distances are equal, the magnitudes of these forces are equal, and they are in opposite directions. Thus, they cancel out: ${\vec{F}}_{O A} + {\vec{F}}_{O C} = 0$ .
Force due to $q_{B}$ and $q_{D}$ :
The charges $q_{B} = - 5 μ$ C and $q_{D} = - 5 μ$ C are equal and are at opposite corners. The force on $q_{O}$ due to $q_{B}$ ( ${\vec{F}}_{O B}$ ) is attractive and directed towards B. The force on $q_{O}$ due to $q_{D}$ ( ${\vec{F}}_{O D}$ ) is attractive and directed towards D. Again, these forces are equal in magnitude and opposite in direction, so they also cancel out: ${\vec{F}}_{O B} + {\vec{F}}_{O D} = 0$ .

The total force on the charge $q_{O}$ at the center is the vector sum of all forces:

{\vec{F}}_{n e t} = ({\vec{F}}_{O A} + {\vec{F}}_{O C}) + ({\vec{F}}_{O B} + {\vec{F}}_{O D}) = 0 + 0 = 0

The net force on the charge at the center is zero.

Exercise 1.7

Question

(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point?

Solution

(a) An electric field line represents the path that a small positive test charge would take if free to move. A sudden break in a field line would mean that the electric field abruptly ceases to exist at that point and then reappears further along. This is physically unrealistic. In a region free of charges, the electric field is continuous. A break would imply there is no force on the test charge at that point, causing its motion to stop, which contradicts the concept of a continuous field originating from a source charge.

(b) The tangent to an electric field line at any point gives the direction of the net electric field at that point. If two field lines were to cross, there would be two different tangents at the point of intersection. This would imply that the electric field has two different directions at the same single point, which is physically impossible. The net electric field at any point in space must have a single, unique direction.

Exercise 1.8

Question

Two point charges $q_{A} = 3 μ$ C and $q_{B} = - 3 μ$ C are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude $1.5 \times 10^{- 9}$ C is placed at this point, what is the force experienced by the test charge?

Solution

(a) Let the midpoint be O. The distance from A to O is $r_{A} = 10$ cm = $0.1$ m, and the distance from B to O is $r_{B} = 10$ cm = $0.1$ m.

Electric field at O due to $q_{A} = 3 μ$ C ( ${\vec{E}}_{A}$ ) is directed away from A (towards B).
Electric field at O due to $q_{B} = - 3 μ$ C ( ${\vec{E}}_{B}$ ) is directed towards B.

Both fields point in the same direction (from A to B). The net field is the sum of their magnitudes.

E_{A} = k \frac{| q_{A} |}{r_{A}^{2}} = (9 \times 10^{9}) \frac{3 \times 10^{- 6}}{(0.1)^{2}} = 2.7 \times 10^{5} N/C

E_{B} = k \frac{| q_{B} |}{r_{B}^{2}} = (9 \times 10^{9}) \frac{3 \times 10^{- 6}}{(0.1)^{2}} = 2.7 \times 10^{5} N/C

E_{n e t} = E_{A} + E_{B} = 5.4 \times 10^{5} N/C, directed from A to B.

(b) A test charge $q_{t e s t} = - 1.5 \times 10^{- 9}$ C is placed at O. The force is $\vec{F} = q_{t e s t} {\vec{E}}_{n e t}$ .

F = (- 1.5 \times 10^{- 9} C) \times (5.4 \times 10^{5} N/C) = - 8.1 \times 10^{- 4} N

The negative sign indicates the force is in the direction opposite to the electric field. Thus, the force is $8.1 \times 10^{- 4}$ N, directed from B to A.

Exercise 1.9

Question

A system has two charges $q_{A} = 2.5 \times 10^{- 7}$ C and $q_{B} = - 2.5 \times 10^{- 7}$ C located at points A: (0, 0, -15 cm) and B: (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

Solution

Total Charge:

The total charge of the system is the algebraic sum of the individual charges.

Q_{t o t a l} = q_{A} + q_{B} = (2.5 \times 10^{- 7} C) + (- 2.5 \times 10^{- 7} C) = 0

Electric Dipole Moment:

The electric dipole moment $\vec{p}$ is given by $\vec{p} = q (2 \vec{a})$ , where $q$ is the magnitude of either charge and $2 \vec{a}$ is the vector from the negative to the positive charge.

$q = 2.5 \times 10^{- 7}$ C.
The negative charge is at B (0, 0, 15 cm) and the positive charge is at A (0, 0, -15 cm).
The vector $2 \vec{a}$ points from B to A. The distance is 30 cm = 0.3 m.
The direction is along the negative z-axis. So, $2 \vec{a} = - 0.3 \hat{k}$ m.

\vec{p} = (2.5 \times 10^{- 7} C) \times (- 0.3 \hat{k} m) = - 7.5 \times 10^{- 8} C \cdot m \hat{k}

The magnitude of the dipole moment is $7.5 \times 10^{- 8}$ C·m, and its direction is along the negative z-axis.

Exercise 1.10

Question

An electric dipole with dipole moment $4 \times 10^{- 9}$ C·m is aligned at 30° with the direction of a uniform electric field of magnitude $5 \times 10^{4}$ N/C. Calculate the magnitude of the torque acting on the dipole.

Solution

The magnitude of the torque $τ$ on a dipole in a uniform electric field is given by:

τ = p E \sin θ

Given:

Dipole moment, $p = 4 \times 10^{- 9}$ C·m
Electric field, $E = 5 \times 10^{4}$ N/C
Angle, $θ = 30^{\circ}$

τ = (4 \times 10^{- 9}) (5 \times 10^{4}) \sin (30^{\circ})

τ = (20 \times 10^{- 5}) \times (0.5) = 10 \times 10^{- 5} = 10^{- 4} N \cdot m

The magnitude of the torque is $10^{- 4}$ N·m.

Exercise 1.11

Question

A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{- 7}$ C.
(a) Estimate the number of electrons transferred (from which to which?).
(b) Is there a transfer of mass from wool to polythene?

Solution

(a) The polythene piece has a negative charge, which means it has gained electrons. Therefore, the electrons were transferred from the wool to the polythene.

Using the quantization of charge, $q = n e$ , we can find the number of electrons, $n$ .

n = \frac{q}{e} = \frac{- 3 \times 10^{- 7} C}{- 1.6 \times 10^{- 19} C/electron} \approx 1.875 \times 10^{12} electrons

(b) Yes, there is a transfer of mass. Each electron has a mass ( $m_{e} = 9.1 \times 10^{- 31}$ kg). The total mass transferred is:

Δ m = n \times m_{e} = (1.875 \times 10^{12}) \times (9.1 \times 10^{- 31} kg) \approx 1.71 \times 10^{- 18} kg

This mass is extremely small and practically negligible, but a transfer of mass does occur.

Exercise 1.12

Question

(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{- 7}$ C? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Solution

(a) Given $q_{A} = q_{B} = 6.5 \times 10^{- 7}$ C and $r = 50$ cm = $0.5$ m.

Using Coulomb's Law:

F = k \frac{q_{A} q_{B}}{r^{2}} = (9 \times 10^{9}) \frac{(6.5 \times 10^{- 7})^{2}}{(0.5)^{2}}

F = (9 \times 10^{9}) \frac{42.25 \times 10^{- 14}}{0.25} \approx 1.52 \times 10^{- 2} N

(b) The new charges are $q_{A}^{'} = q_{B}^{'} = 2 \times (6.5 \times 10^{- 7})$ C. The new distance is $r^{'} = 0.5 / 2 = 0.25$ m.

The new force is:

F^{'} = k \frac{(2 q_{A}) (2 q_{B})}{(r / 2)^{2}} = k \frac{4 q_{A} q_{B}}{r^{2} / 4} = 16 (k \frac{q_{A} q_{B}}{r^{2}}) = 16 F

F^{'} = 16 \times (1.52 \times 10^{- 2} N) \approx 0.243 N

Exercise 1.13

Question

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Solution

The uniform electric field is directed downwards, from the positive top plate to the negative bottom plate.

Signs of the Charges:
Particles are deflected based on the direction of the force they experience. Opposite charges attract.
Particles 1 and 2 are deflected upwards, towards the positive plate. Therefore, particles 1 and 2 must be negatively charged.
Particle 3 is deflected downwards, towards the negative plate. Therefore, particle 3 must be positively charged.
Highest Charge to Mass Ratio (q/m):
The vertical deflection ( $y$ ) is proportional to the vertical acceleration ( $a_{y}$ ). The acceleration is given by $a_{y} = F / m = (q E) / m$ . So, $y \propto \frac{q}{m}$ .
For a given field and initial velocity, the particle with the greatest vertical deflection from its original path will have the highest charge-to-mass ratio.
Observing the tracks, particle 3 shows the maximum vertical deflection. Therefore, particle 3 has the highest charge to mass ratio.

Exercise 1.14

Question

Consider a uniform electric field E = $3 \times 10^{3} \hat{i}$ N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

Solution

The electric field is $\vec{E} = 3 \times 10^{3} \hat{i}$ N/C.
The area of the square is $A = (10 cm)^{2} = (0.1 m)^{2} = 0.01 m^{2}$ .

(a) The plane is parallel to the yz plane. This means its area vector $\vec{A}$ is parallel to the x-axis. So, $\vec{A} = 0.01 \hat{i} m^{2}$ .
The flux $Φ$ is:

Φ = \vec{E} \cdot \vec{A} = (3 \times 10^{3} \hat{i}) \cdot (0.01 \hat{i}) = 30 {N m}^{2} / C

(b) The normal to the plane makes a 60° angle with the x-axis. This is the angle $θ$ between $\vec{E}$ and $\vec{A}$ .
The flux is:

Φ = E A \cos θ = (3 \times 10^{3} N/C) (0.01 m^{2}) \cos (60^{\circ})

Φ = 30 \times (0.5) = 15 {N m}^{2} / C

Exercise 1.15

Question

What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Solution

For any closed surface placed in a uniform electric field, the total electric flux is zero. This is because the number of field lines entering the surface is equal to the number of field lines leaving it. The flux entering is negative, and the flux leaving is positive, and they cancel out perfectly.

Since the cube is a closed surface and the electric field is uniform, the net flux through the cube is zero.

Exercise 1.16

Question

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 10^{3}$ Nm²/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Solution

(a) According to Gauss's Law, the net flux $Φ$ through a closed surface is related to the net charge enclosed $q_{e n c}$ by $Φ = q_{e n c} / ϵ_{0}$ .
Given $Φ = 8.0 \times 10^{3}$ Nm²/C.

q_{e n c} = Φ \times ϵ_{0} = (8.0 \times 10^{3}) \times (8.854 \times 10^{- 12})

q_{e n c} \approx 7.08 \times 10^{- 8} C \approx 0.07 μ C

(b) No. If the net flux were zero, we could only conclude that the net charge inside the box is zero. It is still possible for the box to contain charges, as long as the algebraic sum of those charges is zero. For example, the box could contain an electric dipole or multiple positive and negative charges that cancel each other out.

Exercise 1.17

Question

A point charge +10 $μ$ C is a distance 5 cm directly above the centre of a square of side 10 cm. What is the magnitude of the electric flux through the square?

Solution

We can use the hint and imagine the square as one face of a cube with an edge length of 10 cm. The point charge $q = + 10 μ C$ is placed 5 cm above the center of the square, which means it is located exactly at the geometric center of this imaginary cube.

According to Gauss's Law, the total electric flux ( $Φ_{t o t a l}$ ) through the closed surface of the cube is:

Φ_{t o t a l} = \frac{q_{e n c}}{ϵ_{0}}

Due to the symmetrical placement of the charge at the center, the electric flux is distributed equally through each of the 6 identical faces of the cube.

Therefore, the flux through one face (the square), $Φ_{s q u a r e}$ , is one-sixth of the total flux:

Φ_{s q u a r e} = \frac{1}{6} Φ_{t o t a l} = \frac{1}{6} \frac{q}{ϵ_{0}}

Substituting the values ( $q = 10 \times 10^{- 6}$ C and $ϵ_{0} = 8.854 \times 10^{- 12}$ C²/N·m²):

Φ_{s q u a r e} = \frac{10 \times 10^{- 6}}{6 \times 8.854 \times 10^{- 12}} = \frac{10^{- 5}}{5.3124 \times 10^{- 11}}

Φ_{s q u a r e} \approx 1.88 \times 10^{5} {N m}^{2} /C

The magnitude of the electric flux through the square is approximately 1.88 × 10⁵ N m²/C.

Exercise 1.18

Question

A point charge of 2.0 µC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Solution

According to Gauss's Law, the net electric flux through a closed surface depends only on the net charge enclosed by the surface, not on the shape or size of the surface.
The formula is $Φ = q_{e n c} / ϵ_{0}$ .

Given $q_{e n c} = 2.0 μ C = 2.0 \times 10^{- 6}$ C.

Φ = \frac{2.0 \times 10^{- 6} C}{8.854 \times 10^{- 12} C^{2} / N \cdot m^{2}} \approx 2.26 \times 10^{5} {N m}^{2} / C

The net electric flux through the surface is approximately $2.26 \times 10^{5}$ N m²/C.

Exercise 1.19

Question

A point charge causes an electric flux of $- 1.0 \times 10^{3}$ Nm²/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Solution

(a) According to Gauss's Law, the electric flux through a closed surface depends only on the enclosed charge, not on the size or shape of the surface. Therefore, if the radius of the spherical surface were doubled, the enclosed charge would remain the same, and the flux would also remain unchanged. The flux would still be $- 1.0 \times 10^{3}$ Nm²/C.

(b) Using Gauss's Law, $Φ = q / ϵ_{0}$ , we can find the charge $q$ .

q = Φ \times ϵ_{0} = (- 1.0 \times 10^{3} {Nm}^{2} / C) \times (8.854 \times 10^{- 12} C^{2} / N \cdot m^{2})

q = - 8.854 \times 10^{- 9} C \approx - 8.85 nC

The value of the point charge is approximately -8.85 nC.

Exercise 1.20

Question

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is $1.5 \times 10^{3}$ N/C and points radially inward, what is the net charge on the sphere?

Solution

For a point outside a charged conducting sphere, the electric field is the same as if all the charge were concentrated at its center: $E = k \frac{| q |}{r^{2}}$ .

Given:

Distance from center, $r = 20$ cm = $0.2$ m.
Electric field magnitude, $E = 1.5 \times 10^{3}$ N/C.

The fact that the field points "radially inward" tells us that the charge

q

on the sphere must be negative.

We can find the magnitude of the charge:

| q | = \frac{E r^{2}}{k} = \frac{(1.5 \times 10^{3} N/C) (0.2 m)^{2}}{9 \times 10^{9} {N m}^{2} / C^{2}}

| q | = \frac{(1.5 \times 10^{3}) (0.04)}{9 \times 10^{9}} = \frac{60}{9 \times 10^{9}} \approx 6.67 \times 10^{- 9} C

Since the charge is negative, the net charge on the sphere is approximately -6.67 nC.

Exercise 1.21

Question

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 µC/m². (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Solution

Given diameter = 2.4 m, so radius $R = 1.2$ m.
Surface charge density $σ = 80.0 μ C / m^{2} = 80.0 \times 10^{- 6} C / m^{2}$ .

(a) The charge on the sphere ( $q$ ) is the surface charge density multiplied by the surface area of the sphere ( $A = 4 π R^{2}$ ).

q = σ \times A = (80.0 \times 10^{- 6}) \times (4 π (1.2)^{2})

q \approx (80.0 \times 10^{- 6}) \times (18.1) \approx 1.45 \times 10^{- 3} C

(b) The total electric flux ( $Φ$ ) leaving the surface is given by Gauss's Law:

Φ = \frac{q}{ϵ_{0}} = \frac{1.45 \times 10^{- 3} C}{8.854 \times 10^{- 12} C^{2} / N \cdot m^{2}}

Φ \approx 1.64 \times 10^{8} {N m}^{2} / C

Exercise 1.22

Question

An infinite line charge produces a field of $9 \times 10^{4}$ N/C at a distance of 2 cm. Calculate the linear charge density.

Solution

The electric field ( $E$ ) from an infinite line charge with linear charge density $λ$ at a perpendicular distance $r$ is given by:

E = \frac{λ}{2 π ϵ_{0} r} = \frac{2 k λ}{r}

We can rearrange this to solve for $λ$ . Given:

$E = 9 \times 10^{4}$ N/C
$r = 2$ cm = $0.02$ m

λ = \frac{E r}{2 k} = \frac{(9 \times 10^{4} N/C) (0.02 m)}{2 \times (9 \times 10^{9} {N m}^{2} / C^{2})}

λ = \frac{1800}{18 \times 10^{9}} = 100 \times 10^{- 9} C/m = 10^{- 7} C/m

The linear charge density is $10^{- 7}$ C/m or 0.1 µC/m.

Exercise 1.23

Question

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0 \times 10^{- 22}$ C/m². What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Solution

Let the first plate have $σ = + 17.0 \times 10^{- 22}$ C/m² and the second plate have $- σ$ . The magnitude of the electric field from a single large plate is $E_{p l a t e} = \frac{σ}{2 ϵ_{0}}$ .

(a) Outer region of the first plate (Region I):

The field from the positive plate points left (away). The field from the negative plate points right (towards). They are equal in magnitude and opposite in direction.

E_{I} = E_{(- σ)} - E_{(+ σ)} = \frac{σ}{2 ϵ_{0}} - \frac{σ}{2 ϵ_{0}} = 0

(b) Outer region of the second plate (Region III):

The field from the positive plate points right (away). The field from the negative plate points left (towards). They are equal in magnitude and opposite in direction.

E_{I I I} = E_{(+ σ)} - E_{(- σ)} = \frac{σ}{2 ϵ_{0}} - \frac{σ}{2 ϵ_{0}} = 0

The field from the positive plate points right. The field from the negative plate also points right. The fields add up.

E_{I I} = E_{(+ σ)} + E_{(- σ)} = \frac{σ}{2 ϵ_{0}} + \frac{σ}{2 ϵ_{0}} = \frac{σ}{ϵ_{0}}

Substituting the values:

E_{I I} = \frac{17.0 \times 10^{- 22} {C/m}^{2}}{8.854 \times 10^{- 12} C^{2} / N \cdot m^{2}} \approx 1.92 \times 10^{- 10} N/C

The electric field is 0 in the outer regions and approximately 1.92 × 10⁻¹⁰ N/C between the plates.

Additional Problems

Question 1

Calculate the ratio of the electrostatic to gravitational interaction forces between two electrons, between two protons. At what value of the specific charge q/m of a particle would these forces become equal (in their absolute values) in the case of interaction of identical particles?

Solution

The electrostatic force between two particles with charges $q_{1}$ and $q_{2}$ separated by a distance $r$ is given by Coulomb's Law:

F_{e} = k \frac{q_{1} q_{2}}{r^{2}}

The gravitational force between two particles with masses $m_{1}$ and $m_{2}$ separated by the same distance $r$ is given by Newton's Law of Gravitation:

F_{g} = G \frac{m_{1} m_{2}}{r^{2}}

The constants are:

$k = \frac{1}{4 π ϵ_{0}} \approx 8.99 \times 10^{9} {N m}^{2} / C^{2}$
$G \approx 6.674 \times 10^{- 11} {N m}^{2} / {kg}^{2}$
Charge of an electron/proton (magnitude), $e = 1.602 \times 10^{- 19} C$
Mass of an electron, $m_{e} = 9.109 \times 10^{- 31} kg$
Mass of a proton, $m_{p} = 1.672 \times 10^{- 27} kg$

Part 1: Ratio for two electrons

For two electrons, $q_{1} = q_{2} = e$ (in magnitude) and $m_{1} = m_{2} = m_{e}$ . The ratio of the magnitudes of the forces is:

\frac{| F_{e} |}{| F_{g} |} = \frac{| k e^{2} / r^{2} |}{| G m_{e}^{2} / r^{2} |} = \frac{k e^{2}}{G m_{e}^{2}}

Plugging in the values:

\frac{| F_{e} |}{| F_{g} |} = \frac{(8.99 \times 10^{9}) \times (1.602 \times 10^{- 19})^{2}}{(6.674 \times 10^{- 11}) \times (9.109 \times 10^{- 31})^{2}}

\frac{| F_{e} |}{| F_{g} |} = \frac{2.307 \times 10^{- 28}}{5.536 \times 10^{- 71}} \approx 4.17 \times 10^{42}

Part 2: Ratio for two protons

For two protons, $q_{1} = q_{2} = e$ (in magnitude) and $m_{1} = m_{2} = m_{p}$ . The ratio of the magnitudes of the forces is:

\frac{| F_{e} |}{| F_{g} |} = \frac{k e^{2}}{G m_{p}^{2}}

Plugging in the values:

\frac{| F_{e} |}{| F_{g} |} = \frac{(8.99 \times 10^{9}) \times (1.602 \times 10^{- 19})^{2}}{(6.674 \times 10^{- 11}) \times (1.672 \times 10^{- 27})^{2}}

\frac{| F_{e} |}{| F_{g} |} = \frac{2.307 \times 10^{- 28}}{1.866 \times 10^{- 64}} \approx 1.24 \times 10^{36}

Part 3: Specific charge for equal forces

For the forces to be equal in magnitude for two identical particles of charge $q$ and mass $m$ :

| F_{e} | = | F_{g} | ⟹ k \frac{q^{2}}{r^{2}} = G \frac{m^{2}}{r^{2}}

This simplifies to:

\frac{q^{2}}{m^{2}} = \frac{G}{k}

The specific charge $| q / m |$ is therefore:

| \frac{q}{m} | = \sqrt{\frac{G}{k}}

Plugging in the values:

| \frac{q}{m} | = \sqrt{\frac{6.674 \times 10^{- 11}}{8.99 \times 10^{9}}} = \sqrt{7.424 \times 10^{- 21}} \approx 8.62 \times 10^{- 11} C/kg

Question 2

What would be the interaction force between two copper spheres, each of mass 1 g, separated by the distance 1 m, if the total electronic charge in them differed from the total charge of the nuclei by one per cent?

Solution

First, we need to find the net charge on each copper sphere. Let's find the total positive charge of the nuclei in 1 g of copper.

Given data:

Mass of each sphere, $m_{s p h e r e} = 1 g = 10^{- 3} kg$ .
Distance between spheres, $r = 1 m$ .
Molar mass of Copper (Cu), $M \approx 63.5 g/mol$ .
Atomic number of Copper, $Z = 29$ .
Avogadro's number, $N_{A} \approx 6.022 \times 10^{23} atoms/mol$ .
Elementary charge, $e = 1.602 \times 10^{- 19} C$ .

Step 1: Calculate the number of copper atoms

The number of moles in 1 g of copper is:

n = \frac{mass}{M} = \frac{1 g}{63.5 g/mol} \approx 0.01575 mol

The number of atoms ( $N$ ) in the sphere is:

N = n \times N_{A} = 0.01575 \times (6.022 \times 10^{23}) \approx 9.48 \times 10^{21} atoms

Step 2: Calculate the total positive charge

Each copper nucleus has a charge of $+ Z e$ . The total positive charge from all the nuclei ( $Q_{p o s}$ ) is:

Q_{p o s} = N \times Z \times e = (9.48 \times 10^{21}) \times 29 \times (1.602 \times 10^{- 19} C)

Q_{p o s} \approx 4.40 \times 10^{5} C

Step 3: Calculate the net charge on a sphere

A neutral sphere would have a total electronic charge $Q_{e l e c} = - Q_{p o s}$ . The problem states that the total electronic charge differs from the total charge of the nuclei by 1%. This implies a charge imbalance. The net charge ( $q$ ) on each sphere is 1% of the magnitude of the total positive (or negative) charge.

q = 0.01 \times Q_{p o s} = 0.01 \times (4.40 \times 10^{5} C) = 4.40 \times 10^{3} C

Step 4: Calculate the interaction force

The two spheres have the same mass and charge imbalance, so they have identical charges $q_{1} = q_{2} = q$ . The electrostatic force ( $F_{e}$ ) between them is repulsive:

F_{e} = k \frac{q^{2}}{r^{2}}

Plugging in the values:

F_{e} = (8.99 \times 10^{9} {N m}^{2} / C^{2}) \frac{(4.40 \times 10^{3} C)^{2}}{(1 m)^{2}}

F_{e} = (8.99 \times 10^{9}) \times (1.936 \times 10^{7}) \approx 1.74 \times 10^{17} N

This is an enormous force, highlighting how perfectly balanced the positive and negative charges are in neutral matter.

Question 3

Two small equally charged spheres, each of mass $m$ , are suspended from the same point by silk threads of length $l$ . The distance between the spheres $x ≪ l$ . Find the rate $d q / d t$ with which the charge leaks off each sphere if their approach velocity varies as $v = a / \sqrt{x}$ , where a is a constant.

Solution

Let's analyze the forces acting on one of the spheres in a state of quasi-equilibrium.

Step 1: Force analysis and equilibrium condition

For each sphere, there are three forces acting on it:

Gravitational force, $F_{g} = m g$ , acting vertically downwards.
Electrostatic repulsive force, $F_{e} = k \frac{q^{2}}{x^{2}}$ , acting horizontally.
Tension, $T$ , in the silk thread.

Let $θ$ be the angle the thread makes with the vertical. In equilibrium:

T \sin θ = F_{e} (horizontal component)

T \cos θ = m g (vertical component)

Dividing the two equations, we get:

\tan θ = \frac{F_{e}}{m g} = \frac{k q^{2}}{m g x^{2}}

Step 2: Small angle approximation

Since the distance between the spheres $x$ is much smaller than the length of the threads $l$ ( $x ≪ l$ ), the angle $θ$ is small. We can use the small angle approximation:

\tan θ \approx \sin θ = \frac{x / 2}{l} = \frac{x}{2 l}

Step 3: Relate charge $q$ to distance $x$

Equating the two expressions for $\tan θ$ :

\frac{x}{2 l} = \frac{k q^{2}}{m g x^{2}}

Now, we solve for $q^{2}$ in terms of $x$ :

q^{2} = \frac{m g}{2 l k} x^{3}

Taking the square root to find $q$ :

q = \sqrt{\frac{m g}{2 l k}} x^{3 / 2}

Step 4: Differentiate with respect to time

We need to find the rate of charge leakage, $d q / d t$ . We use the chain rule:

\frac{d q}{d t} = \frac{d q}{d x} \frac{d x}{d t}

First, differentiate $q$ with respect to $x$ :

\frac{d q}{d x} = \sqrt{\frac{m g}{2 l k}} \cdot \frac{3}{2} x^{(3 / 2 - 1)} = \frac{3}{2} \sqrt{\frac{m g}{2 l k}} x^{1 / 2}

Step 5: Determine $d x / d t$

The problem states that the approach velocity of each sphere is $v = a / \sqrt{x}$ . Since both spheres are moving towards each other with this speed, the rate at which the distance $x$ between them decreases is the sum of their speeds:

| \frac{d x}{d t} | = v + v = 2 v = \frac{2 a}{\sqrt{x}}

Since the charge is leaking, $q$ decreases, the repulsive force $F_{e}$ decreases, and the spheres move closer. Thus, $x$ is decreasing, which means $d x / d t$ is negative. Also, $q$ is decreasing, so $d q / d t$ is negative.

\frac{d x}{d t} = - \frac{2 a}{\sqrt{x}}

Step 6: Calculate $d q / d t$

Now, we substitute the expressions for $d q / d x$ and $d x / d t$ into the chain rule equation:

\frac{d q}{d t} = (\frac{3}{2} \sqrt{\frac{m g}{2 l k}} x^{1 / 2}) (- \frac{2 a}{\sqrt{x}})

The terms involving $x$ cancel out:

\frac{d q}{d t} = - \frac{3}{2} \cdot 2 a \cdot \sqrt{\frac{m g}{2 l k}} = - 3 a \sqrt{\frac{m g}{2 l k}}

The question asks for the rate with which the charge leaks off, which is a positive quantity representing the magnitude of the rate of change. Let this rate be $R = | d q / d t |$ .

R = | \frac{d q}{d t} | = 3 a \sqrt{\frac{m g}{2 l k}}

Substituting $k = 1 / (4 π ϵ_{0})$ gives:

R = 3 a \sqrt{\frac{m g (4 π ϵ_{0})}{2 l}} = 3 a \sqrt{\frac{2 π ϵ_{0} m g}{l}}

Question 4

Two positive charges $q_{1}$ and $q_{2}$ are located at the points with radius vector $r_{1}$ and $r_{2}$ . Find a negative charge $q_{3}$ and a radius vector $r_{3}$ of the point at which it has to be placed for the force acting on each of the three charges to be equal to zero.

Solution

For the net force on all three charges to be zero, two conditions must be met:

The three charges must be collinear.
The forces on each charge from the other two must cancel out.

Since $q_{1}$ and $q_{2}$ are positive, they repel each other. For a third charge $q_{3}$ to be in equilibrium, it must be placed on the line connecting $q_{1}$ and $q_{2}$ . Furthermore, for $q_{1}$ and $q_{2}$ to be in equilibrium, $q_{3}$ must be a negative charge placed between them to provide an attractive force that balances their mutual repulsion.

Step 1: Find the position $r_{3}$ of charge $q_{3}$

Let $q_{3}$ be placed at a distance $x$ from $q_{1}$ along the line segment connecting $q_{1}$ and $q_{2}$ . The distance between $q_{1}$ and $q_{2}$ is $d = | r_{2} - r_{1} |$ . The distance from $q_{3}$ to $q_{2}$ is then $d - x$ .

For the net force on $q_{3}$ to be zero, the attractive forces from $q_{1}$ and $q_{2}$ must be equal in magnitude:

F_{31} = F_{32} ⟹ k \frac{q_{1} | q_{3} |}{x^{2}} = k \frac{q_{2} | q_{3} |}{(d - x)^{2}}

\frac{q_{1}}{x^{2}} = \frac{q_{2}}{(d - x)^{2}}

Taking the square root of both sides (since distances are positive):

\frac{\sqrt{q_{1}}}{x} = \frac{\sqrt{q_{2}}}{d - x} ⟹ \sqrt{q_{1}} (d - x) = x \sqrt{q_{2}}

Solving for $x$ :

d \sqrt{q_{1}} - x \sqrt{q_{1}} = x \sqrt{q_{2}} ⟹ d \sqrt{q_{1}} = x (\sqrt{q_{1}} + \sqrt{q_{2}})

x = d \frac{\sqrt{q_{1}}}{\sqrt{q_{1}} + \sqrt{q_{2}}}

The radius vector $r_{3}$ lies on the line segment between $r_{1}$ and $r_{2}$ . It can be expressed as a linear combination:

r_{3} = r_{1} + \frac{x}{d} (r_{2} - r_{1}) = (1 - \frac{x}{d}) r_{1} + \frac{x}{d} r_{2}

Substituting the ratio $x / d$ :

r_{3} = (1 - \frac{\sqrt{q_{1}}}{\sqrt{q_{1}} + \sqrt{q_{2}}}) r_{1} + (\frac{\sqrt{q_{1}}}{\sqrt{q_{1}} + \sqrt{q_{2}}}) r_{2}

r_{3} = (\frac{\sqrt{q_{2}}}{\sqrt{q_{1}} + \sqrt{q_{2}}}) r_{1} + (\frac{\sqrt{q_{1}}}{\sqrt{q_{1}} + \sqrt{q_{2}}}) r_{2} = \frac{\sqrt{q_{2}} r_{1} + \sqrt{q_{1}} r_{2}}{\sqrt{q_{1}} + \sqrt{q_{2}}}

Step 2: Find the magnitude of charge $q_{3}$

For the entire system to be in equilibrium, the net force on charge $q_{1}$ (and $q_{2}$ ) must also be zero. The repulsive force from $q_{2}$ must be balanced by the attractive force from $q_{3}$ .

F_{12} = F_{13} ⟹ k \frac{q_{1} q_{2}}{d^{2}} = k \frac{q_{1} | q_{3} |}{x^{2}}

\frac{q_{2}}{d^{2}} = \frac{| q_{3} |}{x^{2}} ⟹ | q_{3} | = q_{2} {(\frac{x}{d})}^{2}

Substitute the ratio $x / d$ we found earlier:

| q_{3} | = q_{2} {(\frac{\sqrt{q_{1}}}{\sqrt{q_{1}} + \sqrt{q_{2}}})}^{2} = \frac{q_{1} q_{2}}{(\sqrt{q_{1}} + \sqrt{q_{2}})^{2}}

Since we established that $q_{3}$ must be negative:

q_{3} = - \frac{q_{1} q_{2}}{(\sqrt{q_{1}} + \sqrt{q_{2}})^{2}}

Question 5

A thin wire ring of radius $R$ has an electric charge $q$ . What will be the increment of the force stretching the wire if a point charge $q_{0}$ is placed at the ring's centre?

Solution

The "force stretching the wire" refers to the tension ( $T$ ) within the wire ring. The problem asks for the increment in this tension when a charge $q_{0}$ is added at the center. This increment is the tension caused solely by the interaction between the charge $q_{0}$ at the center and the charge $q$ distributed on the ring.

Step 1: Force on a small element of the ring

Consider a small element of the ring with length $d l$ . This element subtends an angle $d θ = d l / R$ at the center. The charge on this element, $d q$ , is:

d q = λ d l

where $λ = q / (2 π R)$ is the linear charge density of the ring.

The charge $q_{0}$ at the center exerts a radially outward force, $d F_{o u t}$ , on this element $d q$ .

d F_{o u t} = k \frac{q_{0} d q}{R^{2}} = k \frac{q_{0} (λ d l)}{R^{2}}

Step 2: Balancing force from tension

This outward radial force is balanced by the net inward component of the tension forces ( $T$ ) acting at the ends of the element $d l$ . The two tension forces are tangential to the ring. The angle between them is $d θ$ . The net inward radial force they produce is:

d F_{i n} = 2 T \sin (\frac{d θ}{2})

For a very small element, $d θ$ is small, so we can use the approximation $\sin (x) \approx x$ .

d F_{i n} \approx 2 T (\frac{d θ}{2}) = T d θ

Since $d l = R d θ$ , we have $d θ = d l / R$ . So,

d F_{i n} = T \frac{d l}{R}

Step 3: Equating forces to find the tension

In equilibrium, the outward force from the central charge is balanced by the inward pull from the tension.

d F_{o u t} = d F_{i n}

k \frac{q_{0} (λ d l)}{R^{2}} = T \frac{d l}{R}

We can cancel $d l$ from both sides and solve for the tension $T$ . This tension is the increment, $Δ T$ , caused by $q_{0}$ .

Δ T = T = k \frac{q_{0} λ}{R}

Now, substitute the expression for the linear charge density $λ = q / (2 π R)$ :

Δ T = k \frac{q_{0}}{R} (\frac{q}{2 π R}) = \frac{k q q_{0}}{2 π R^{2}}

Using the constant $k = 1 / (4 π ϵ_{0})$ , the expression becomes:

Δ T = \frac{1}{4 π ϵ_{0}} \frac{q q_{0}}{2 π R^{2}} = \frac{q q_{0}}{8 π^{2} ϵ_{0} R^{2}}

This is the increment of the force stretching the wire.

Question 6

A positive point charge 50 µC is located in the plane xy at the point with radius vector $r_{0} = 2 i + 3 j$ , where i and j are the unit vector of the x and y axis. Find the vector of the force acting on the charge if the vector of electric field strength $E = 8 i - 5 j$ . Here $r_{0}$ and r are expressed in metres.

Solution

The force $F$ experienced by a point charge $q$ placed in an electric field $E$ is given by the formula:

F = q E

Given data:

Charge, $q = 50 μ C = 50 \times 10^{- 6} C$ .
Electric field vector, $E = (8 i - 5 j)$ N/C.

The position of the charge, $r_{0} = 2 i + 3 j$ , is given. However, the electric field $E$ is given as a constant vector, which means the field is uniform. In a uniform field, the force on the charge is the same regardless of its position. Therefore, the position vector $r_{0}$ is not needed to calculate the force.

Calculation of the Force Vector

We substitute the given values of $q$ and $E$ into the force equation:

F = (50 \times 10^{- 6} C) \times (8 i - 5 j N/C)

F = (50 \times 10^{- 6} \times 8) i - (50 \times 10^{- 6} \times 5) j

F = (400 \times 10^{- 6}) i - (250 \times 10^{- 6}) j N

F = (4.0 \times 10^{- 4} i - 2.5 \times 10^{- 4} j) N

This can also be expressed in millinewtons (mN):

F = (0.40 i - 0.25 j) mN

Question 7

Point charges $q$ and $- q$ are located at the vertices of a square with diagonals $2 l$ as shown in Fig. 3.1. Find the magnitude of the electric field strength at a point located symmetrically with respect to the vertices of the square at a distance $x$ from its centre.

Solution

Let's set up a coordinate system to solve this problem. Let the centre of the square be the origin (0, 0, 0). The square lies in the $x y$ -plane, and the point of interest, P, is on the $z$ -axis at $(0, 0, x)$ .

Step 1: Determine the coordinates of the charges

The diagonals of the square have length $2 l$ , which means the distance from the center to each vertex is $l$ . Based on the provided figure (Fig. 3.1), the charges are placed such that the top two vertices have charge $+ q$ and the bottom two have charge $- q$ . We can assign coordinates as follows, aligning the y-axis with the vertical axis of the figure:

$+ q$ at $(- d, d, 0)$ and $(d, d, 0)$
$- q$ at $(- d, - d, 0)$ and $(d, - d, 0)$

Here, the distance from the center to a vertex is $l = \sqrt{d^{2} + d^{2}} = d \sqrt{2}$ . Thus, $d = l / \sqrt{2}$ . The coordinates are:

$A (+ q) : (- l / \sqrt{2}, l / \sqrt{2}, 0)$
$B (+ q) : (l / \sqrt{2}, l / \sqrt{2}, 0)$
$C (- q) : (- l / \sqrt{2}, - l / \sqrt{2}, 0)$
$D (- q) : (l / \sqrt{2}, - l / \sqrt{2}, 0)$

Step 2: Calculate the electric field vector

The distance $r$ from any vertex to point P(0, 0, x) is the same:

r = \sqrt{(l / \sqrt{2})^{2} + (l / \sqrt{2})^{2} + x^{2}} = \sqrt{l^{2} + x^{2}}

The electric field vector from a point charge $q_{i}$ at $r_{i}^{'}$ to a point $r$ is $E_{i} = k \frac{q_{i} (r - r_{i}^{'})}{| r - r_{i}^{'} |^{3}}$ . We sum the vectors from all four charges:

E_{t o t a l} = \frac{k}{r^{3}} [q_{A} (r - r_{A}^{'}) + q_{B} (r - r_{B}^{'}) + q_{C} (r - r_{C}^{'}) + q_{D} (r - r_{D}^{'})]

E_{t o t a l} = \frac{k}{r^{3}} [q (l / \sqrt{2}, - l / \sqrt{2}, x) + q (- l / \sqrt{2}, - l / \sqrt{2}, x) - q (l / \sqrt{2}, l / \sqrt{2}, x) - q (- l / \sqrt{2}, l / \sqrt{2}, x)]

Now, we sum the components:

x-component: $k q / r^{3} \times [l / \sqrt{2} - l / \sqrt{2} - l / \sqrt{2} + l / \sqrt{2}] = 0$
y-component: $k q / r^{3} \times [- l / \sqrt{2} - l / \sqrt{2} - l / \sqrt{2} - l / \sqrt{2}] = - k q / r^{3} \times (4 l / \sqrt{2}) = - 2 \sqrt{2} k q l / r^{3}$
z-component: $k q / r^{3} \times [x + x - x - x] = 0$

The total electric field vector is:

E_{t o t a l} = - \frac{2 \sqrt{2} k q l}{r^{3}} j

The direction is along the negative y-axis, from the pair of positive charges towards the pair of negative charges, which is physically consistent.

Step 3: Find the magnitude

The magnitude of the electric field is the magnitude of this vector.

E = | E_{t o t a l} | = \frac{2 \sqrt{2} k q l}{r^{3}}

Substituting $r = \sqrt{l^{2} + x^{2}}$ , we get:

E = \frac{2 \sqrt{2} k q l}{(l^{2} + x^{2})^{3 / 2}}

Using $k = 1 / (4 π ϵ_{0})$ :

E = \frac{2 \sqrt{2} q l}{4 π ϵ_{0} (l^{2} + x^{2})^{3 / 2}} = \frac{\sqrt{2} q l}{2 π ϵ_{0} (l^{2} + x^{2})^{3 / 2}}

Question 8

A thin half-ring of radius $R = 20 cm$ is uniformly charged with a total charge $q = 0.70 nC$ . Find the magnitude of the electric field strength at the curvature centre of this half-ring.

Solution

Let the half-ring be in the $x y$ -plane, centered at the origin $(0, 0)$ . Due to symmetry, the electric field components along the x-axis will cancel out. Only the component along the y-axis (or axis perpendicular to the diameter connecting the ends of the half-ring) will remain.

Let the half-ring extend from angle $θ = 0$ to $θ = π$ . The linear charge density $λ$ is the total charge $q$ divided by the length of the half-ring, which is $π R$ :

λ = \frac{q}{π R}

Consider a small element of the half-ring with length $d l = R d θ$ . The charge on this element is $d q = λ d l = λ R d θ$ .

The electric field $d E$ due to this element $d q$ at the center is:

d E = k \frac{d q}{R^{2}} = k \frac{λ R d θ}{R^{2}} = k \frac{λ d θ}{R}

Let's consider the components of $d E$ . By symmetry, if the half-ring is symmetric about the y-axis (i.e., from $- π / 2$ to $π / 2$ ), then the x-components will cancel. If it's symmetric about the x-axis (from $0$ to $π$ ), then the y-components will sum up.

Assume the half-ring spans from $θ = 0$ to $θ = π$ as shown in the typical setup (e.g., semicircle in the upper half-plane). The coordinates of $d q$ are $(R \cos θ, R \sin θ)$ . The field $d E$ points from $d q$ to the origin, or from the origin to $d q$ depending on the setup. Let's assume the question means "electric field at the center of curvature" meaning the origin for a half-ring in the upper plane.

The electric field vector $d E$ from $d q$ at $(R \cos θ, R \sin θ)$ to the origin $(0, 0)$ has components:

d E_{x} = d E \cos (π - θ) = - d E \cos θ

d E_{y} = d E \sin (π - θ) = d E \sin θ

Integrating from $θ = 0$ to $θ = π$ :

E_{x} = \int_{0}^{π} - k \frac{λ}{R} \cos θ d θ = - k \frac{λ}{R} [\sin θ]_{0}^{π} = 0

E_{y} = \int_{0}^{π} k \frac{λ}{R} \sin θ d θ = k \frac{λ}{R} [- \cos θ]_{0}^{π} = k \frac{λ}{R} (- \cos π + \cos 0) = k \frac{λ}{R} (1 + 1) = 2 k \frac{λ}{R}

So the net electric field is purely in the y-direction.

E = E_{y} = 2 k \frac{λ}{R}

Substitute $λ = q / (π R)$ :

E = 2 k \frac{q}{π R^{2}}

Given values:

$R = 20 cm = 0.20 m$
$q = 0.70 nC = 0.70 \times 10^{- 9} C$
$k = 8.99 \times 10^{9} {N m}^{2} / C^{2}$

Calculate $E$ :

E = \frac{2 \times (8.99 \times 10^{9} {N m}^{2} / C^{2}) \times (0.70 \times 10^{- 9} C)}{π \times (0.20 m)^{2}}

E = \frac{12.586}{π \times 0.04} = \frac{12.586}{0.12566} \approx 100.15 N/C

Thus, the magnitude of the electric field strength at the curvature center of the half-ring is approximately $100 N/C$ .

Question 9

A thin wire ring of radius $r$ carries a charge $q$ . Find the magnitude of the electric field strength on the axis of the ring as a function of distance $l$ from its centre. Investigate the obtained function at $l ≫ r$ . Find the maximum strength magnitude and the corresponding distance $l$ . Draw the approximate plot of the function $E (l)$ .

Solution

Part 1: Electric field strength on the axis

Consider a ring of radius $r$ with total charge $q$ distributed uniformly. Let the axis of the ring be the $z$ -axis, and the center of the ring be the origin. We want to find the electric field at a point P on the axis at a distance $l$ from the center (i.e., at $(0, 0, l)$ ).

Consider a small charge element $d q$ on the ring. The distance from $d q$ to point P is $s = \sqrt{r^{2} + l^{2}}$ . The electric field $d E$ due to $d q$ is:

d E = k \frac{d q}{s^{2}} = k \frac{d q}{r^{2} + l^{2}}

Due to symmetry, the components of $d E$ perpendicular to the axis (i.e., in the $x y$ -plane) will cancel out when integrated over the entire ring. Only the axial component (along the $z$ -axis) will contribute.

The axial component $d E_{z}$ is $d E \cos ϕ$ , where $ϕ$ is the angle between the vector $s$ and the axis. From the geometry, $\cos ϕ = l / s = l / \sqrt{r^{2} + l^{2}}$ .

d E_{z} = d E \cos ϕ = k \frac{d q}{r^{2} + l^{2}} \frac{l}{\sqrt{r^{2} + l^{2}}} = k \frac{l d q}{(r^{2} + l^{2})^{3 / 2}}

To find the total electric field $E$ , we integrate $d E_{z}$ over the entire ring. Since $k$ , $l$ , $r$ are constants for this integration, they can be pulled out:

E = \int d E_{z} = \frac{k l}{(r^{2} + l^{2})^{3 / 2}} \int d q

The integral $\int d q$ is the total charge $q$ on the ring.

E (l) = \frac{k q l}{(r^{2} + l^{2})^{3 / 2}}

Part 2: Investigation at $l ≫ r$

When $l ≫ r$ , we can approximate $(r^{2} + l^{2})^{3 / 2} \approx (l^{2})^{3 / 2} = l^{3}$ .

E (l) \approx \frac{k q l}{l^{3}} = \frac{k q}{l^{2}}

This result is expected: at distances far from the ring, the ring effectively behaves like a point charge $q$ located at its center, producing an electric field that falls off as $1 / l^{2}$ .

Part 3: Maximum strength magnitude

To find the maximum strength, we need to find the value of $l$ for which $d E / d l = 0$ .

E (l) = k q l (r^{2} + l^{2})^{- 3 / 2}

Using the product rule and chain rule for differentiation:

\frac{d E}{d l} = k q [(r^{2} + l^{2})^{- 3 / 2} + l (- \frac{3}{2}) (r^{2} + l^{2})^{- 5 / 2} (2 l)]

\frac{d E}{d l} = k q [(r^{2} + l^{2})^{- 3 / 2} - 3 l^{2} (r^{2} + l^{2})^{- 5 / 2}]

Set $d E / d l = 0$ :

(r^{2} + l^{2})^{- 3 / 2} = 3 l^{2} (r^{2} + l^{2})^{- 5 / 2}

Multiply both sides by $(r^{2} + l^{2})^{5 / 2}$ :

(r^{2} + l^{2}) = 3 l^{2}

r^{2} = 2 l^{2}

l^{2} = \frac{r^{2}}{2}

l = \frac{r}{\sqrt{2}}

Since $l$ is a distance, it must be positive. The electric field strength is maximum at $l = r / \sqrt{2}$ .

Now, substitute this value of $l$ back into the expression for $E (l)$ to find the maximum strength:

E_{m a x} = \frac{k q (r / \sqrt{2})}{{(r^{2} + (r / \sqrt{2})^{2})}^{3 / 2}} = \frac{k q r / \sqrt{2}}{{(r^{2} + r^{2} / 2)}^{3 / 2}}

E_{m a x} = \frac{k q r / \sqrt{2}}{{(3 r^{2} / 2)}^{3 / 2}} = \frac{k q r / \sqrt{2}}{(3 / 2)^{3 / 2} r^{3}} = \frac{k q r / \sqrt{2}}{(3 \sqrt{3} / (2 \sqrt{2})) r^{3}}

E_{m a x} = \frac{k q}{(3 \sqrt{3} / 2) r^{2}} = \frac{2 k q}{3 \sqrt{3} r^{2}}

Part 4: Plot of $E (l)$

The function $E (l)$ starts at $E (0) = 0$ (due to symmetry, field at center is zero for a uniformly charged ring). It increases to a maximum value at $l = r / \sqrt{2}$ , and then decreases, approaching zero as $1 / l^{2}$ for large $l$ . The plot would look like a bell curve, but only for $l \geq 0$ .

At $l = 0$ , $E (0) = 0$ .
At $l = r / \sqrt{2}$ , $E (l)$ is maximum.
As $l \to \infty$ , $E (l) \to 0$ as $k q / l^{2}$ .

The general shape is an increase from zero, reaching a peak, then a decay back to zero. The curve is symmetric about $l = 0$ if we consider both positive and negative $l$ .

Approximate Plot:

(A plot with E(l) on the y-axis and l on the x-axis, showing a curve starting at 0, increasing to a peak at $l = r / \sqrt{2}$ , then decreasing asymptotically to 0.)

Question 10

A point charge $q$ is located at the centre of a thin ring of radius $R$ with uniformly distributed charge $- q$ . Find the magnitude of the electric field strength vector at the point lying on the axis of the ring at a distance $x$ from its centre, if $x ≫ R$ .

Solution

The system consists of two parts contributing to the electric field:

A point charge $q$ at the center of the ring.
A uniformly charged ring of radius $R$ with total charge $- q$ .

Let the center of the ring be the origin and the axis of the ring be the $z$ -axis. We want to find the electric field at a point P $(0, 0, x)$ on the axis.

Step 1: Electric field due to the point charge $q$

The electric field $E_{1}$ due to a point charge $q$ at the origin, at a distance $x$ along the z-axis, is given by Coulomb's law:

E_{1} = k \frac{q}{x^{2}} \hat{k}

where $\hat{k}$ is the unit vector along the z-axis. The magnitude is $E_{1} = k | q | / x^{2}$ .

Step 2: Electric field due to the charged ring $- q$

From the previous problem (Question 9), the electric field $E_{2}$ on the axis of a uniformly charged ring of radius $R$ and charge $Q_{r i n g}$ at a distance $x$ from its center is:

E_{2} = \frac{k Q_{r i n g} x}{(R^{2} + x^{2})^{3 / 2}} \hat{k}

In this problem, $Q_{r i n g} = - q$ . So,

E_{2} = \frac{k (- q) x}{(R^{2} + x^{2})^{3 / 2}} \hat{k} = - \frac{k q x}{(R^{2} + x^{2})^{3 / 2}} \hat{k}

Step 3: Total electric field

The total electric field $E_{t o t a l}$ is the vector sum of $E_{1}$ and $E_{2}$ :

E_{t o t a l} = E_{1} + E_{2} = (k \frac{q}{x^{2}} - \frac{k q x}{(R^{2} + x^{2})^{3 / 2}}) \hat{k}

E_{t o t a l} = k q (\frac{1}{x^{2}} - \frac{x}{(R^{2} + x^{2})^{3 / 2}}) \hat{k}

Step 4: Approximation for $x ≫ R$

When $x ≫ R$ , we can use the binomial approximation $(1 + u)^{n} \approx 1 + n u$ for small $u$ . Rewrite the term $(R^{2} + x^{2})^{- 3 / 2}$ :

(R^{2} + x^{2})^{- 3 / 2} = (x^{2} (1 + R^{2} / x^{2}))^{- 3 / 2} = (x^{2})^{- 3 / 2} (1 + R^{2} / x^{2})^{- 3 / 2}

= \frac{1}{x^{3}} (1 - \frac{3}{2} \frac{R^{2}}{x^{2}} + \dots)

Now substitute this back into the expression for $E_{t o t a l}$ :

E_{t o t a l} = k q (\frac{1}{x^{2}} - x \cdot \frac{1}{x^{3}} (1 - \frac{3}{2} \frac{R^{2}}{x^{2}})) \hat{k}

E_{t o t a l} = k q (\frac{1}{x^{2}} - \frac{1}{x^{2}} (1 - \frac{3}{2} \frac{R^{2}}{x^{2}})) \hat{k}

E_{t o t a l} = k q (\frac{1}{x^{2}} - \frac{1}{x^{2}} + \frac{3}{2} \frac{R^{2}}{x^{4}}) \hat{k}

E_{t o t a l} = \frac{3 k q R^{2}}{2 x^{4}} \hat{k}

The magnitude of the electric field strength vector is:

E_{t o t a l} = \frac{3 k q R^{2}}{2 x^{4}}

This result shows that when $x ≫ R$ , the electric field falls off as $1 / x^{4}$ . This is characteristic of a dipole field. The total charge of the system is $q + (- q) = 0$ , so it's a dipole. The dipole moment of the ring and point charge system needs to be considered for a full dipole analysis, but the resulting field dependence ( $1 / x^{4}$ ) is consistent with a dipole field at large distances.

Question 11

A system consists of a thin charged wire ring of radius $R$ and a very long uniformly charged thread oriented along the axis of the ring, with one of its ends coinciding with the centre of the ring. The total charge of the ring is equal to $q$ . The charge of the thread (per unit length) is equal to $λ$ . Find the interaction force between the ring and the thread.

Solution

Let the ring be in the $x y$ -plane with its center at the origin. The long uniformly charged thread is along the $z$ -axis, starting from the origin (center of the ring) and extending to positive infinity. The ring has total charge $q$ , and the thread has linear charge density $λ$ .

We need to find the interaction force. We can calculate the force exerted by the ring on the thread, or by the thread on the ring. It is usually easier to calculate the force exerted by the continuous charge distribution (ring) on a small element of the other (thread), and then integrate, or vice versa.

Let's find the force exerted by the ring on an infinitesimal segment $d z$ of the thread located at a distance $z$ from the center of the ring. The charge on this segment is $d q_{t h r e a d} = λ d z$ .

Step 1: Electric field due to the ring on its axis

The electric field produced by a uniformly charged ring of radius $R$ and total charge $q$ at a point on its axis at a distance $z$ from its center is given by:

E_{r i n g} (z) = \frac{k q z}{(R^{2} + z^{2})^{3 / 2}}

This field is directed along the z-axis (assuming $q$ is positive). So $E_{r i n g} (z) = E_{r i n g} (z) \hat{k}$ .

Step 2: Force on the segment $d q_{t h r e a d}$

The force $d F$ on the charge $d q_{t h r e a d}$ at position $z$ is $d F = d q_{t h r e a d} E_{r i n g} (z)$ .

d F = (λ d z) \frac{k q z}{(R^{2} + z^{2})^{3 / 2}}

This force is directed along the z-axis. So the total force $F$ on the thread is found by integrating $d F$ from $z = 0$ (center of the ring) to $z = \infty$ (end of the thread).

F = \int_{0}^{\infty} \frac{k q λ z}{(R^{2} + z^{2})^{3 / 2}} d z

Step 3: Evaluate the integral

Let $u = R^{2} + z^{2}$ . Then $d u = 2 z d z$ , so $z d z = \frac{1}{2} d u$ . When $z = 0$ , $u = R^{2}$ . When $z = \infty$ , $u = \infty$ .

F = k q λ \int_{R^{2}}^{\infty} u^{- 3 / 2} \frac{1}{2} d u

F = \frac{k q λ}{2} {[\frac{u^{- 1 / 2}}{- 1 / 2}]}_{R^{2}}^{\infty}

F = \frac{k q λ}{2} {[- 2 u^{- 1 / 2}]}_{R^{2}}^{\infty}

F = - k q λ {[\frac{1}{\sqrt{u}}]}_{R^{2}}^{\infty}

F = - k q λ (0 - \frac{1}{\sqrt{R^{2}}})

F = k q λ \frac{1}{R}

Using $k = 1 / (4 π ϵ_{0})$ , the force is:

F = \frac{q λ}{4 π ϵ_{0} R}

The direction of the force is along the axis of the ring, specifically outwards if $q$ and $λ$ have the same sign (repulsive), or inwards if they have opposite signs (attractive).

Question 12

A thin nonconducting ring of radius $R$ has a linear charge density $λ = λ_{0} \cos ϕ$ , where $λ_{0}$ is a constant, $ϕ$ is the azimuthal angle. Find the magnitude of the electric field strength

(a) at the centre of the ring;

(b) on the axis of the ring as a function of the distance $x$ from its centre. Investigate the obtained function at $x ≫ R$ .

Solution

The linear charge density is given by $λ = λ_{0} \cos ϕ$ . The charge element $d q$ on a small arc length $d l = R d ϕ$ is:

d q = λ d l = (λ_{0} \cos ϕ) R d ϕ

Part (a): Electric field strength at the center of the ring

Let the center of the ring be the origin $(0, 0)$ . Consider a charge element $d q$ at an angle $ϕ$ . Its coordinates are $(R \cos ϕ, R \sin ϕ)$ . The electric field $d E$ it produces at the origin is directed from $d q$ to the origin, so it points towards $(- R \cos ϕ, - R \sin ϕ)$ .

The magnitude of $d E$ is $d E = k \frac{d q}{R^{2}} = k \frac{λ_{0} \cos ϕ R d ϕ}{R^{2}} = k \frac{λ_{0} \cos ϕ}{R} d ϕ$ .

The components of $d E$ are:

d E_{x} = d E \cos (π + ϕ) = - d E \cos ϕ = - k \frac{λ_{0} \cos^{2} ϕ}{R} d ϕ

d E_{y} = d E \sin (π + ϕ) = - d E \sin ϕ = - k \frac{λ_{0} \cos ϕ \sin ϕ}{R} d ϕ

To find the total electric field $E$ at the center, we integrate over the entire ring (from $ϕ = 0$ to $2 π$ ):

E_{x} = \int_{0}^{2 π} - k \frac{λ_{0} \cos^{2} ϕ}{R} d ϕ = - \frac{k λ_{0}}{R} \int_{0}^{2 π} \frac{1 + \cos (2 ϕ)}{2} d ϕ

E_{x} = - \frac{k λ_{0}}{2 R} {[ϕ + \frac{1}{2} \sin (2 ϕ)]}_{0}^{2 π} = - \frac{k λ_{0}}{2 R} (2 π) = - \frac{π k λ_{0}}{R}

E_{y} = \int_{0}^{2 π} - k \frac{λ_{0} \cos ϕ \sin ϕ}{R} d ϕ = - \frac{k λ_{0}}{R} \int_{0}^{2 π} \frac{1}{2} \sin (2 ϕ) d ϕ

E_{y} = - \frac{k λ_{0}}{2 R} {[- \frac{1}{2} \cos (2 ϕ)]}_{0}^{2 π} = - \frac{k λ_{0}}{2 R} (- \frac{1}{2} (1 - 1)) = 0

So, the electric field at the center of the ring is $E = - \frac{π k λ_{0}}{R} \hat{i}$ .

The magnitude is $E = \frac{π k λ_{0}}{R}$ . Using $k = 1 / (4 π ϵ_{0})$ :

E = \frac{π λ_{0}}{4 π ϵ_{0} R} = \frac{λ_{0}}{4 ϵ_{0} R}

Part (b): Electric field strength on the axis as a function of distance $x$

Let the point be $(0, 0, x)$ on the axis. The distance from a charge element $d q$ at $(R \cos ϕ, R \sin ϕ, 0)$ to the point $(0, 0, x)$ is $s = \sqrt{R^{2} + x^{2}}$ .

The electric field $d E$ due to $d q$ has an axial component $d E_{z}$ and a radial component $d E_{r}$ . The magnitude of $d E$ is $d E = k \frac{d q}{s^{2}} = k \frac{λ_{0} \cos ϕ R d ϕ}{R^{2} + x^{2}}$ .

The radial component $d E_{r}$ is directed towards the center of the ring in the $x y$ -plane (perpendicular to the axis). Its components are:

d E_{x} = d E_{r} \cos ϕ = \frac{k d q}{s^{2}} \frac{R}{s} \cos ϕ = \frac{k (λ_{0} \cos ϕ R d ϕ)}{(R^{2} + x^{2})} \frac{R}{\sqrt{R^{2} + x^{2}}} \cos ϕ = \frac{k λ_{0} R^{2} \cos^{2} ϕ}{(R^{2} + x^{2})^{3 / 2}} d ϕ

d E_{y} = d E_{r} \sin ϕ = \frac{k d q}{s^{2}} \frac{R}{s} \sin ϕ = \frac{k λ_{0} R^{2} \cos ϕ \sin ϕ}{(R^{2} + x^{2})^{3 / 2}} d ϕ

The axial component $d E_{z}$ is:

d E_{z} = d E \cos θ = k \frac{d q}{s^{2}} \frac{x}{s} = \frac{k λ_{0} \cos ϕ R d ϕ}{(R^{2} + x^{2})} \frac{x}{\sqrt{R^{2} + x^{2}}} = \frac{k λ_{0} R x \cos ϕ}{(R^{2} + x^{2})^{3 / 2}} d ϕ

Now, integrate over $ϕ$ from $0$ to $2 π$ :

E_{x} = \int_{0}^{2 π} \frac{k λ_{0} R^{2} \cos^{2} ϕ}{(R^{2} + x^{2})^{3 / 2}} d ϕ = \frac{k λ_{0} R^{2}}{(R^{2} + x^{2})^{3 / 2}} \int_{0}^{2 π} \cos^{2} ϕ d ϕ

We know $\int_{0}^{2 π} \cos^{2} ϕ d ϕ = π$ .

E_{x} = \frac{k λ_{0} π R^{2}}{(R^{2} + x^{2})^{3 / 2}}

E_{y} = \int_{0}^{2 π} \frac{k λ_{0} R^{2} \cos ϕ \sin ϕ}{(R^{2} + x^{2})^{3 / 2}} d ϕ = \frac{k λ_{0} R^{2}}{(R^{2} + x^{2})^{3 / 2}} \int_{0}^{2 π} \frac{1}{2} \sin (2 ϕ) d ϕ = 0

E_{z} = \int_{0}^{2 π} \frac{k λ_{0} R x \cos ϕ}{(R^{2} + x^{2})^{3 / 2}} d ϕ = \frac{k λ_{0} R x}{(R^{2} + x^{2})^{3 / 2}} \int_{0}^{2 π} \cos ϕ d ϕ = 0

So, the total electric field vector is $E = E_{x} \hat{i}$ . The magnitude is:

E (x) = \frac{k π λ_{0} R^{2}}{(R^{2} + x^{2})^{3 / 2}}

Using $k = 1 / (4 π ϵ_{0})$ :

E (x) = \frac{π λ_{0} R^{2}}{4 π ϵ_{0} (R^{2} + x^{2})^{3 / 2}} = \frac{λ_{0} R^{2}}{4 ϵ_{0} (R^{2} + x^{2})^{3 / 2}}

Investigation at $x ≫ R$

When $x ≫ R$ , we can approximate $(R^{2} + x^{2})^{3 / 2} \approx (x^{2})^{3 / 2} = x^{3}$ .

E (x) \approx \frac{λ_{0} R^{2}}{4 ϵ_{0} x^{3}}

This result shows that for large distances along the axis, the electric field falls off as $1 / x^{3}$ . This is characteristic of a dipole field. The charge distribution $λ = λ_{0} \cos ϕ$ corresponds to an electric dipole. The total charge of the ring is $\int_{0}^{2 π} λ_{0} \cos ϕ R d ϕ = λ_{0} R [\sin ϕ]_{0}^{2 π} = 0$ . So it is a neutral system with a dipole moment. The dipole moment for such a distribution lies in the plane of the ring and points along the direction where $\cos ϕ$ is maximum (e.g., $x$ -axis if $ϕ$ is measured from $x$ -axis).

Question 13

A thin straight rod of length $2 a$ carrying a uniformly distributed charge $q$ is located in vacuum. Find the magnitude of the electric field strength as a function of the distance $r$ from the rod's centre along the straight line.

(a) perpendicular to the rod and passing through its centre;

(b) coinciding with the rod's direction (at the points lying outside the rod).

Investigate the obtained expression at $r ≫ a$ .

Solution

Let the rod be placed along the x-axis, centered at the origin, extending from $x = - a$ to $x = + a$ . The total length is $2 a$ . The uniform linear charge density is $λ = q / (2 a)$ . The electrostatic constant is $k = 1 / (4 π ϵ_{0})$ .

Part (a): Electric field on the perpendicular bisector

Consider a point P at $(0, r, 0)$ on the perpendicular bisector (y-axis). An infinitesimal charge element $d q = λ d x$ is located at $(x, 0, 0)$ . The distance from $d q$ to P is $s = \sqrt{x^{2} + r^{2}}$ . The electric field $d E$ due to $d q$ points from $d q$ to P. Its magnitude is $d E = k \frac{d q}{s^{2}} = k \frac{λ d x}{x^{2} + r^{2}}$ .

Due to symmetry, the x-components of $d E$ will cancel out over the entire rod. Only the y-component ( $d E_{y}$ ) will contribute. The y-component is $d E_{y} = d E \cos α$ , where $α$ is the angle between the vector from $d q$ to P and the y-axis. From the geometry, $\cos α = r / s = r / \sqrt{x^{2} + r^{2}}$ .

d E_{y} = k \frac{λ d x}{x^{2} + r^{2}} \frac{r}{\sqrt{x^{2} + r^{2}}} = k λ r \frac{d x}{(x^{2} + r^{2})^{3 / 2}}

Integrate $d E_{y}$ from $x = - a$ to $x = + a$ :

E = \int_{- a}^{a} k λ r \frac{d x}{(x^{2} + r^{2})^{3 / 2}} = k λ r {[\frac{x}{r^{2} \sqrt{x^{2} + r^{2}}}]}_{- a}^{a}

E = \frac{k λ}{r} (\frac{a}{\sqrt{a^{2} + r^{2}}} - \frac{- a}{\sqrt{(- a)^{2} + r^{2}}}) = \frac{k λ}{r} (\frac{2 a}{\sqrt{a^{2} + r^{2}}})

Substitute $λ = q / (2 a)$ :

E = \frac{k (q / (2 a))}{r} \frac{2 a}{\sqrt{a^{2} + r^{2}}} = \frac{k q}{r \sqrt{a^{2} + r^{2}}}

The direction of $E$ is along the y-axis (perpendicular to the rod, away from the rod if $q$ is positive).

Part (b): Electric field on the line coinciding with the rod's direction (axial line)

Consider a point P at $(r_{p}, 0, 0)$ on the x-axis, where $r_{p} > a$ (outside the rod). An infinitesimal charge element $d q = λ d x$ is located at $(x, 0, 0)$ . The distance from $d q$ to P is $s = | r_{p} - x |$ . All $d E$ contributions are along the x-axis. If $q > 0$ , the field points away from the rod.

E = \int_{- a}^{a} k \frac{d q}{(r_{p} - x)^{2}} = \int_{- a}^{a} k \frac{λ d x}{(r_{p} - x)^{2}}

Integrate:

E = k λ {[\frac{1}{r_{p} - x}]}_{- a}^{a} = k λ (\frac{1}{r_{p} - a} - \frac{1}{r_{p} - (- a)})

E = k λ (\frac{1}{r_{p} - a} - \frac{1}{r_{p} + a}) = k λ (\frac{(r_{p} + a) - (r_{p} - a)}{(r_{p} - a) (r_{p} + a)}) = k λ \frac{2 a}{r_{p}^{2} - a^{2}}

Substitute $λ = q / (2 a)$ :

E = k \frac{q}{2 a} \frac{2 a}{r_{p}^{2} - a^{2}} = \frac{k q}{r_{p}^{2} - a^{2}}

The direction of $E$ is along the x-axis, away from the rod if $q$ is positive.

Investigation at $r ≫ a$ (or $r_{p} ≫ a$ )

For part (a), $E = \frac{k q}{r \sqrt{a^{2} + r^{2}}}$ . If $r ≫ a$ , then $\sqrt{a^{2} + r^{2}} \approx \sqrt{r^{2}} = r$ .

E \approx \frac{k q}{r \cdot r} = \frac{k q}{r^{2}}

For part (b), $E = \frac{k q}{r_{p}^{2} - a^{2}}$ . If $r_{p} ≫ a$ , then $r_{p}^{2} - a^{2} \approx r_{p}^{2}$ .

E \approx \frac{k q}{r_{p}^{2}}

In both cases, at large distances, the electric field of the rod approaches that of a point charge $q$ located at the origin. This is expected, as from far away, the dimensions of the rod become negligible compared to the distance to the observation point.

Question 14

A very long straight uniformly charged thread carries a charge $λ$ per unit length. Find the magnitude and direction of the electric field strength at a point which is at a distance $y$ from the thread and lies on the perpendicular passing through one of the thread's ends.

Solution

Let the thread be placed along the x-axis, starting from the origin $(0, 0, 0)$ and extending to positive infinity ( $x \geq 0$ ). Let the point P be at $(0, y, 0)$ on the y-axis, at a distance $y$ from the thread, and perpendicular to one of its ends (the origin). The electrostatic constant is $k = 1 / (4 π ϵ_{0})$ .

Consider an infinitesimal charge element $d q = λ d x$ located at $(x, 0, 0)$ . The position vector of P is $r_{P} = (0, y, 0)$ . The position vector of $d q$ is $r_{d q} = (x, 0, 0)$ . The vector from $d q$ to P is $r = r_{P} - r_{d q} = (- x, y, 0)$ . The distance from $d q$ to P is $s = \sqrt{(- x)^{2} + y^{2}} = \sqrt{x^{2} + y^{2}}$ .

The electric field $d E$ due to $d q$ is:

d E = k \frac{d q}{s^{3}} r = k \frac{λ d x}{(x^{2} + y^{2})^{3 / 2}} (- x \hat{i} + y \hat{j})

We need to find the components $E_{x}$ and $E_{y}$ by integrating over $x$ from $0$ to $\infty$ :

E_{x} = \int_{0}^{\infty} k λ \frac{- x d x}{(x^{2} + y^{2})^{3 / 2}} = - k λ \int_{0}^{\infty} \frac{x d x}{(x^{2} + y^{2})^{3 / 2}}

Let $u = x^{2} + y^{2}$ , so $d u = 2 x d x$ . When $x = 0$ , $u = y^{2}$ . When $x = \infty$ , $u = \infty$ .

E_{x} = - k λ \int_{y^{2}}^{\infty} \frac{1}{2} u^{- 3 / 2} d u = - k λ {[\frac{1}{2} \frac{u^{- 1 / 2}}{- 1 / 2}]}_{y^{2}}^{\infty} = - k λ {[- \frac{1}{\sqrt{u}}]}_{y^{2}}^{\infty}

E_{x} = - k λ (0 - (- \frac{1}{\sqrt{y^{2}}})) = - k λ (\frac{1}{| y |}) = - \frac{k λ}{y} (assuming y > 0)

For the y-component:

E_{y} = \int_{0}^{\infty} k λ \frac{y d x}{(x^{2} + y^{2})^{3 / 2}} = k λ y \int_{0}^{\infty} \frac{d x}{(x^{2} + y^{2})^{3 / 2}}

Using the standard integral $\int \frac{d x}{(x^{2} + A^{2})^{3 / 2}} = \frac{x}{A^{2} \sqrt{x^{2} + A^{2}}}$ (here $A = y$ ):

E_{y} = k λ y {[\frac{x}{y^{2} \sqrt{x^{2} + y^{2}}}]}_{0}^{\infty} = \frac{k λ}{y} {[\frac{x}{\sqrt{x^{2} + y^{2}}}]}_{0}^{\infty}

E_{y} = \frac{k λ}{y} (lim_{x \to \infty} \frac{x}{x \sqrt{1 + y^{2} / x^{2}}} - 0) = \frac{k λ}{y} (1) = \frac{k λ}{y}

The electric field vector is $E = - \frac{k λ}{y} \hat{i} + \frac{k λ}{y} \hat{j}$ .

Magnitude of the electric field strength

E = | E | = \sqrt{E_{x}^{2} + E_{y}^{2}} = \sqrt{{(- \frac{k λ}{y})}^{2} + {(\frac{k λ}{y})}^{2}} = \sqrt{2 \frac{k^{2} λ^{2}}{y^{2}}} = \frac{k λ \sqrt{2}}{| y |}

Using $k = 1 / (4 π ϵ_{0})$ :

E = \frac{\sqrt{2} λ}{4 π ϵ_{0} y}

Direction of the electric field strength

The field components are $E_{x} = - \frac{k λ}{y}$ and $E_{y} = \frac{k λ}{y}$ . The angle $θ$ that the resultant vector makes with the positive x-axis is:

\tan θ = \frac{E_{y}}{E_{x}} = \frac{k λ / y}{- k λ / y} = - 1

Since $E_{x}$ is negative and $E_{y}$ is positive, the vector lies in the second quadrant. Therefore, $θ = 135^{\circ}$ (or $3 π / 4$ radians) with respect to the positive x-axis. This means the electric field vector is directed at $45^{\circ}$ with respect to the thread, pointing towards the starting end of the thread and away from it perpendicularly.

Question 15

A thread carrying a uniform charge $λ$ per unit length has the configurations shown in Fig. 3.2a and b. Assuming a curvature radius $R$ to be considerably less than the length of the thread, find the magnitude of the electric field strength at the point $O$ .

Solution

Let $O$ be the origin $(0, 0)$ . The uniform linear charge density is $λ$ . The electrostatic constant is $k = 1 / (4 π ϵ_{0})$ .

(a) Configuration (a)

This configuration consists of two parts:

A semi-infinite straight wire along the positive x-axis starting at $(R, 0)$ and extending to positive infinity.
A quarter-circle of radius $R$ in the fourth quadrant, connecting $(R, 0)$ to $(0, - R)$ .

The electric field at $O$ is the vector sum of fields from these two parts.

1. Electric field from the semi-infinite straight wire:

The wire extends from $(R, 0)$ to $(\infty, 0)$ . Consider an element $d x^{'}$ at $(x^{'}, 0)$ where $x^{'} \geq R$ . The vector from $d x^{'}$ to the origin $(0, 0)$ is $(- x^{'}, 0)$ . The magnitude of $d E$ is $k \frac{λ d x^{'}}{x^{' 2}}$ . Its direction is along the negative x-axis.

E_{s t r a i g h t} = \int_{R}^{\infty} k \frac{λ d x^{'}}{x^{' 2}} (- \hat{i}) = - k λ {[- \frac{1}{x^{'}}]}_{R}^{\infty} \hat{i} = - k λ (0 - (- \frac{1}{R})) \hat{i} = - \frac{k λ}{R} \hat{i}

2. Electric field from the quarter-circle:

The quarter-circle extends from $ϕ = 0$ (point $(R, 0)$ ) to $ϕ = - π / 2$ (point $(0, - R)$ ). An element $d q = λ d l = λ R d ϕ$ is at $(R \cos ϕ, R \sin ϕ)$ . The vector from $d q$ to the origin is $(- R \cos ϕ, - R \sin ϕ)$ . The magnitude of $d E$ is $k \frac{d q}{R^{2}} = k \frac{λ R d ϕ}{R^{2}} = k \frac{λ}{R} d ϕ$ .

The components of $d E$ are:

d E_{x} = d E \cos (π + ϕ) = - d E \cos ϕ = - \frac{k λ}{R} \cos ϕ d ϕ

d E_{y} = d E \sin (π + ϕ) = - d E \sin ϕ = - \frac{k λ}{R} \sin ϕ d ϕ

Integrate from $ϕ = 0$ to $ϕ = - π / 2$ :

E_{a r c, x} = \int_{0}^{- π / 2} - \frac{k λ}{R} \cos ϕ d ϕ = - \frac{k λ}{R} [\sin ϕ]_{0}^{- π / 2} = - \frac{k λ}{R} (- 1 - 0) = \frac{k λ}{R}

E_{a r c, y} = \int_{0}^{- π / 2} - \frac{k λ}{R} \sin ϕ d ϕ = - \frac{k λ}{R} [- \cos ϕ]_{0}^{- π / 2} = - \frac{k λ}{R} (0 - (- 1)) = \frac{k λ}{R}

So, $E_{a r c} = \frac{k λ}{R} \hat{i} + \frac{k λ}{R} \hat{j}$ .

The total electric field at O for configuration (a) is the vector sum:

E_{a} = E_{s t r a i g h t} + E_{a r c} = (- \frac{k λ}{R} \hat{i}) + (\frac{k λ}{R} \hat{i} + \frac{k λ}{R} \hat{j}) = \frac{k λ}{R} \hat{j}

The magnitude is $E_{a} = \frac{k λ}{R}$ . Using $k = 1 / (4 π ϵ_{0})$ , $E_{a} = \frac{λ}{4 π ϵ_{0} R}$ .

(b) Configuration (b)

This configuration consists of three parts:

A semi-infinite straight wire along the positive x-axis from $(R, 0)$ to positive infinity.
A semi-infinite straight wire along the negative x-axis from $(- R, 0)$ to negative infinity.
A semicircle of radius $R$ in the upper half-plane, connecting $(- R, 0)$ to $(R, 0)$ .

1. Electric field from the two semi-infinite straight wires:

Right wire (from $(R, 0)$ to $(\infty, 0)$ ): From part (a), it contributes $- \frac{k λ}{R} \hat{i}$ .
Left wire (from $(- \infty, 0)$ to $(- R, 0)$ ): Consider an element $d x^{'}$ at $(x^{'}, 0)$ where $x^{'} \leq - R$ . The vector from $d x^{'}$ to the origin is $(- x^{'}, 0)$ . Since $x^{'}$ is negative, $- x^{'}$ is positive, so the field points along the positive x-axis.
$E_{s t r a i g h t 2} = \int_{- \infty}^{- R} k \frac{λ d x^{'}}{x^{' 2}} (\hat{i}) = k λ {[- \frac{1}{x^{'}}]}_{- \infty}^{- R} \hat{i} = k λ (- \frac{1}{- R} - 0) \hat{i} = \frac{k λ}{R} \hat{i}$

The total contribution from the two straight wires is $E_{s t r a i g h t 1} + E_{s t r a i g h t 2} = (- \frac{k λ}{R} \hat{i}) + (\frac{k λ}{R} \hat{i}) = 0$ .

2. Electric field from the semi-circle:

The semi-circle extends from $ϕ = π$ (point $(- R, 0)$ ) to $ϕ = 0$ (point $(R, 0)$ ). An element $d q = λ R d ϕ$ is at $(R \cos ϕ, R \sin ϕ)$ . The vector from $d q$ to the origin is $(- R \cos ϕ, - R \sin ϕ)$ .

E_{a r c, x} = \int_{π}^{0} - \frac{k λ}{R} \cos ϕ d ϕ = - \frac{k λ}{R} [\sin ϕ]_{π}^{0} = 0

E_{a r c, y} = \int_{π}^{0} - \frac{k λ}{R} \sin ϕ d ϕ = - \frac{k λ}{R} [- \cos ϕ]_{π}^{0} = - \frac{k λ}{R} (- \cos (0) - (- \cos (π))) = - \frac{k λ}{R} (- 1 - (- 1)) = - \frac{2 k λ}{R}

So, $E_{a r c} = - \frac{2 k λ}{R} \hat{j}$ .

The total electric field at O for configuration (b) is the vector sum:

E_{b} = 0 + (- \frac{2 k λ}{R} \hat{j}) = - \frac{2 k λ}{R} \hat{j}

The magnitude is $E_{b} = \frac{2 k λ}{R}$ . Using $k = 1 / (4 π ϵ_{0})$ , $E_{b} = \frac{2 λ}{4 π ϵ_{0} R} = \frac{λ}{2 π ϵ_{0} R}$ .

Question 16

A sphere of radius $r$ carries a surface charge of density $σ = a \cdot r$ , where $a$ is a constant vector, and $r$ is the radius vector of a point of the sphere relative to its centre. Find the electric field strength vector at the centre of the sphere.

Solution

Let the center of the sphere be the origin $(0, 0, 0)$ . The radius of the sphere is $R$ (using $R$ instead of $r$ to avoid confusion with the position vector $r$ ). An infinitesimal surface element $d A$ at a position $r$ on the sphere has charge $d q = σ d A = (a \cdot r) d A$ . The electric field $d E$ at the origin due to this charge element $d q$ is given by:

d E = k \frac{d q}{R^{2}} (- \frac{r}{R}) = - k \frac{(a \cdot r) d A}{R^{3}} r

To find the total electric field $E$ at the center, we integrate $d E$ over the entire surface of the sphere:

E = \int_{S} - k \frac{(a \cdot r) r}{R^{3}} d A = - \frac{k}{R^{3}} \int_{S} (a \cdot r) r d A

Let $a = (a_{x}, a_{y}, a_{z})$ and $r = (x, y, z)$ . Then $a \cdot r = a_{x} x + a_{y} y + a_{z} z$ . The integral $\int_{S} (a \cdot r) r d A$ is a vector integral. Let's consider its x-component:

{(\int_{S} (a \cdot r) r d A)}_{x} = \int_{S} (a_{x} x + a_{y} y + a_{z} z) x d A

= a_{x} \int_{S} x^{2} d A + a_{y} \int_{S} x y d A + a_{z} \int_{S} x z d A

For a sphere centered at the origin, by symmetry, the integrals of products of distinct coordinates are zero: $\int_{S} x y d A = 0$ , $\int_{S} x z d A = 0$ , $\int_{S} y z d A = 0$ .

Also, by symmetry, $\int_{S} x^{2} d A = \int_{S} y^{2} d A = \int_{S} z^{2} d A$ . The sum of these integrals is $\int_{S} (x^{2} + y^{2} + z^{2}) d A = \int_{S} R^{2} d A = R^{2} \int_{S} d A = R^{2} (4 π R^{2}) = 4 π R^{4}$ . Therefore, $\int_{S} x^{2} d A = \frac{1}{3} (4 π R^{4})$ .

So, the x-component of the integral is:

{(\int_{S} (a \cdot r) r d A)}_{x} = a_{x} \frac{4 π R^{4}}{3}

Similarly for the y and z components:

{(\int_{S} (a \cdot r) r d A)}_{y} = a_{y} \frac{4 π R^{4}}{3}

{(\int_{S} (a \cdot r) r d A)}_{z} = a_{z} \frac{4 π R^{4}}{3}

Thus, the vector integral is:

\int_{S} (a \cdot r) r d A = \frac{4 π R^{4}}{3} (a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k}) = \frac{4 π R^{4}}{3} a

Substitute this back into the expression for $E$ :

E = - \frac{k}{R^{3}} (\frac{4 π R^{4}}{3} a) = - \frac{4 π k R}{3} a

Using $k = 1 / (4 π ϵ_{0})$ :

E = - \frac{4 π}{3} \frac{1}{4 π ϵ_{0}} R a = - \frac{R}{3 ϵ_{0}} a

The electric field strength vector at the center of the sphere is proportional to the constant vector $a$ and points in the opposite direction.

Question 17

Suppose the surface charge density over a sphere of radius $R$ depends on a polar angle $θ$ as $σ = σ_{0} \cos θ$ , where $σ_{0}$ is a positive constant. Show that such a charge distribution can be represented as a result of a small relative shift of two uniformly charged balls of radius $R$ whose charges are equal in magnitude and opposite in sign. Resorting to this representation, find the electric field strength vector inside the given sphere.

Solution

Part 1: Representation as shifted uniformly charged balls

Consider two uniformly charged balls (spheres) of radius $R$ . Let ball 1 have uniform volume charge density $+ ρ$ and be centered at $d / 2$ . Let ball 2 have uniform volume charge density $- ρ$ and be centered at $- d / 2$ . Assume the small shift $d$ is along the z-axis, so $d = (0, 0, d)$ , with $d ≪ R$ . The composite system (two overlapping spheres) can be viewed as having a net volume charge density. However, for an observation point outside the shifted spheres or far from them, it resembles a dipole. The problem focuses on the surface charge density of the *original* sphere, formed by the overlap of these two. For a small displacement $d$ along the z-axis, the excess charge on the surface of the sphere of radius $R$ (centered at the origin) will be approximately $σ \approx ρ d \cos θ$ . This approximation arises because the point at $(R, θ, ϕ)$ is closer to the center of the positive sphere by $(d / 2) \cos θ$ and farther from the center of the negative sphere by $(d / 2) \cos θ$ . The imbalance of the uniform volume charge inside results in an effective surface charge density. Comparing this to the given $σ = σ_{0} \cos θ$ , we can identify $ρ d = σ_{0}$ . Thus, the surface charge distribution $σ = σ_{0} \cos θ$ can be represented as the superposition of two slightly shifted uniformly charged balls of radius $R$ and opposite uniform volume charge densities $\pm ρ$ , where $ρ = σ_{0} / d$ . The axis of the shift is the same as the axis defining the polar angle $θ$ (the z-axis).

Part 2: Electric field strength vector inside the given sphere

The electric field inside a uniformly charged sphere of radius $R$ with volume charge density $ρ$ (at a point $r^{'}$ relative to its center) is given by Gauss's Law as $E^{'} = \frac{ρ r^{'}}{3 ϵ_{0}}$ .

For the composite system inside the sphere of radius $R$ (at a point $r$ relative to the origin):

Ball 1 (charge $+ ρ$ ) is centered at $r_{1} = d / 2$ . The field it produces at $r$ is $E_{1} = \frac{ρ (r - d / 2)}{3 ϵ_{0}}$ .
Ball 2 (charge $- ρ$ ) is centered at $r_{2} = - d / 2$ . The field it produces at $r$ is $E_{2} = \frac{- ρ (r - (- d / 2))}{3 ϵ_{0}} = - \frac{ρ (r + d / 2)}{3 ϵ_{0}}$ .

The total electric field inside the sphere is the superposition of these two fields:

E_{t o t a l} = E_{1} + E_{2} = \frac{ρ}{3 ϵ_{0}} (r - d / 2 - (r + d / 2))

E_{t o t a l} = \frac{ρ}{3 ϵ_{0}} (- d) = - \frac{ρ d}{3 ϵ_{0}}

This shows that the electric field inside is uniform. Since we found that $ρ d = σ_{0}$ (where $d$ is the magnitude of $d$ ), and $d = d \hat{k}$ (assuming $θ$ is the polar angle w.r.t. z-axis), we can substitute $ρ = σ_{0} / d$ into the expression:

E_{t o t a l} = - \frac{(σ_{0} / d) (d \hat{k})}{3 ϵ_{0}} = - \frac{σ_{0}}{3 ϵ_{0}} \hat{k}

The magnitude of the electric field strength vector inside the sphere is:

E = \frac{σ_{0}}{3 ϵ_{0}}

The field is uniform and points along the negative z-axis (opposite to the direction where $\cos θ$ is positive and decreasing, consistent with dipole fields).

Question 18

Find the electric field strength vector at the centre of a ball of radius $R$ with volume charge density $ρ = a \cdot r$ , where $a$ is a constant vector, and $r$ is a radius vector drawn from the ball's centre.

Solution

Let the center of the ball be the origin $(0, 0, 0)$ . The volume charge density is $ρ = a \cdot r$ . We want to find the electric field $E$ at the origin. The electrostatic constant is $k = 1 / (4 π ϵ_{0})$ .

Consider an infinitesimal volume element $d V^{'}$ at position $r^{'}$ relative to the center. The charge in this element is $d q = ρ (r^{'}) d V^{'} = (a \cdot r^{'}) d V^{'}$ . The electric field $d E$ at the origin due to this charge element $d q$ is given by:

d E = k \frac{d q}{(r^{'})^{2}} (- \frac{r^{'}}{r^{'}}) = - k \frac{(a \cdot r^{'}) r^{'}}{(r^{'})^{3}} d V^{'}

To find the total electric field $E$ at the center, we integrate $d E$ over the entire volume of the ball (from $r^{'} = 0$ to $R$ ):

E = \int_{V} - k \frac{(a \cdot r^{'}) r^{'}}{(r^{'})^{3}} d V^{'}

We use spherical coordinates for integration. $r^{'} = (r^{'} \sin θ \cos ϕ, r^{'} \sin θ \sin ϕ, r^{'} \cos θ)$ , and $d V^{'} = (r^{'})^{2} \sin θ d r^{'} d θ d ϕ$ . Let $a$ be aligned with the z-axis, so $a = a_{z} \hat{k}$ . Then $a \cdot r^{'} = a_{z} r^{'} \cos θ$ . Due to symmetry, the x and y components of $E$ will cancel out. Only the z-component might be non-zero. Let's calculate $E_{z}$ .

E_{z} = \int_{0}^{R} \int_{0}^{π} \int_{0}^{2 π} - k \frac{(a_{z} r^{'} \cos θ) (r^{'} \cos θ)}{(r^{'})^{3}} (r^{'})^{2} \sin θ d r^{'} d θ d ϕ

Simplifying the terms involving $r^{'}$ and trigonometric functions:

E_{z} = - k a_{z} \int_{0}^{R} (r^{'})^{2} d r^{'} \int_{0}^{π} \cos^{2} θ \sin θ d θ \int_{0}^{2 π} d ϕ

Evaluate each integral separately:

Radial integral: $\int_{0}^{R} (r^{'})^{2} d r^{'} = {[\frac{(r^{'})^{3}}{3}]}_{0}^{R} = \frac{R^{3}}{3}$ .
Polar angle integral: $\int_{0}^{π} \cos^{2} θ \sin θ d θ$ . Let $u = \cos θ$ , so $d u = - \sin θ d θ$ . When $θ = 0$ , $u = 1$ . When $θ = π$ , $u = - 1$ .
$\int_{1}^{- 1} - u^{2} d u = \int_{- 1}^{1} u^{2} d u = {[\frac{u^{3}}{3}]}_{- 1}^{1} = \frac{1^{3}}{3} - \frac{(- 1)^{3}}{3} = \frac{1}{3} - (- \frac{1}{3}) = \frac{2}{3}$
Azimuthal angle integral: $\int_{0}^{2 π} d ϕ = 2 π$ .

Combine these results for $E_{z}$ :

E_{z} = - k a_{z} (\frac{R^{3}}{3}) (\frac{2}{3}) (2 π) = - k a_{z} \frac{4 π R^{3}}{9}

Since we aligned $a$ with the z-axis, $a_{z} = | a |$ . The result is therefore general for any direction of $a$ : the electric field vector $E$ is parallel to $a$ but in the opposite direction.

E = - k \frac{4 π R^{3}}{9} a

Using $k = 1 / (4 π ϵ_{0})$ :

E = - \frac{1}{4 π ϵ_{0}} \frac{4 π R^{3}}{9} a = - \frac{R^{3}}{9 ϵ_{0}} a

The electric field strength vector at the center of the ball is proportional to $a$ and points in the opposite direction.

Question 19

A very long uniformly charged thread oriented along the axis of a circle of radius $R$ rests on its centre with one of the ends. The charge of the thread per unit length is equal to $λ$ . Find the flux of the vector $E$ across the circle area.

Solution

Let the circle be in the $x y$ -plane, centered at the origin $(0, 0, 0)$ . The thread is along the z-axis, starting from the origin ( $z = 0$ ) and extending to positive infinity ( $z \geq 0$ ). The linear charge density of the thread is $λ$ . We need to find the electric flux $Φ_{E}$ through the area of the circle. The electrostatic constant is $k = 1 / (4 π ϵ_{0})$ .

The electric flux is given by $Φ_{E} = \int E \cdot d A$ . The area vector for the circle is $d A = d A \hat{k}$ (where $\hat{k}$ is the unit vector along the positive z-axis), and $d A = r^{'} d r^{'} d ϕ$ in cylindrical coordinates (where $r^{'}$ is the radial distance from the center in the $x y$ -plane, and $ϕ$ is the azimuthal angle).

First, let's find the electric field $E$ at a point $(r^{'}, 0, 0)$ on the circle (using $r^{'}$ as the radial coordinate within the circle, ranging from $0$ to $R$ ). An infinitesimal charge element $d q = λ d z$ is located at $(0, 0, z)$ on the thread. The vector from $d q$ to the point $(r^{'}, 0, 0)$ on the circle is $r_{p o i n t} - r_{d q} = r^{'} \hat{i} - z \hat{k}$ . The distance from $d q$ to the point is $s = \sqrt{(r^{'})^{2} + z^{2}}$ .

The electric field $d E$ due to $d q$ is:

d E = k \frac{d q}{s^{3}} (r^{'} \hat{i} - z \hat{k}) = k \frac{λ d z}{((r^{'})^{2} + z^{2})^{3 / 2}} (r^{'} \hat{i} - z \hat{k})

We are interested in the component of $E$ perpendicular to the circle's plane (i.e., the z-component, $E_{z}$ ), since $d A$ is along $\hat{k}$ .

E_{z} = \int_{0}^{\infty} k λ \frac{- z d z}{((r^{'})^{2} + z^{2})^{3 / 2}}

Let $u = (r^{'})^{2} + z^{2}$ , so $d u = 2 z d z$ . When $z = 0$ , $u = (r^{'})^{2}$ . When $z = \infty$ , $u = \infty$ .

E_{z} = k λ \int_{(r^{'})^{2}}^{\infty} \frac{- 1}{2} u^{- 3 / 2} d u = - k λ {[\frac{1}{2} \frac{u^{- 1 / 2}}{- 1 / 2}]}_{(r^{'})^{2}}^{\infty} = - k λ {[- \frac{1}{\sqrt{u}}]}_{(r^{'})^{2}}^{\infty}

E_{z} = - k λ (0 - (- \frac{1}{\sqrt{(r^{'})^{2}}})) = - k λ (\frac{1}{| r^{'} |}) = - \frac{k λ}{r^{'}}

The electric field component perpendicular to the circle is thus $E_{z} = - k λ / r^{'}$ . This component points towards the negative z-axis.

Now, calculate the flux $Φ_{E}$ through the circle area of radius $R$ . The elementary area is $d A = r^{'} d r^{'} d ϕ$ , and the area vector is $d A = r^{'} d r^{'} d ϕ \hat{k}$ .

Φ_{E} = \int_{A} E \cdot d A = \int_{0}^{2 π} \int_{0}^{R} E_{z} (r^{'} d r^{'} d ϕ)

Φ_{E} = \int_{0}^{2 π} \int_{0}^{R} (- \frac{k λ}{r^{'}}) (r^{'} d r^{'} d ϕ)

Φ_{E} = \int_{0}^{2 π} \int_{0}^{R} - k λ d r^{'} d ϕ

Φ_{E} = - k λ (\int_{0}^{R} d r^{'}) (\int_{0}^{2 π} d ϕ)

Φ_{E} = - k λ (R) (2 π) = - 2 π k λ R

The magnitude of the flux is $2 π k λ R$ . Using $k = 1 / (4 π ϵ_{0})$ :

Φ_{E} = - 2 π (\frac{1}{4 π ϵ_{0}}) λ R = - \frac{λ R}{2 ϵ_{0}}

The negative sign indicates that the flux is inwards (towards the negative z-axis) if $λ$ is positive. The question asks for the flux of the vector $E$ across the circle area, which implies the signed flux. If the magnitude is required, it is $\frac{λ R}{2 ϵ_{0}}$ .