JEE Advanced PYQ Physics – Electrostatic Potential and Equipotential Surfaces Questions

Q 1 :

An electric dipole is formed by two charges $+ q$ and $- q$ located in the $x y$ -plane at (0, 2) mm and $(0, - 2)$ mm, respectively, as shown in the figure. The electric potential at point $P (100, 100) mm$ due to the dipole is $V_{0}$ . The charges $+ q$ and $- q$ are then moved to the points $(- 1, 2) mm$ and $(1, - 2) mm$ , respectively. What is the value of electric potential at $P$ due to the new dipole? [2023]

[IMAGE 703]

$V_{0} / 4$
$V_{0} / 2$
$V_{0} / \sqrt{2}$
$3 V_{0} / 4$

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(2)

[IMAGE 704]

Since electric potential due to a dipole at a general point is given by

$V = \frac{K p \cos θ}{r^{3}}$

$V_{1} \propto \frac{p_{1} \cos θ_{1}}{r_{1}^{3}}$

Electric potential at point $P (100, 100)$ $V_{1} = V_{0}$ , and the electric potential at the new point is

$V_{2} \propto \frac{p_{2} \cos θ_{2}}{r_{2}^{3}}$

$\frac{V_{2}}{V_{1}} = \frac{p_{2} \cos θ_{2}}{p_{1} \cos θ_{1}} = \frac{q (- 2 \hat{i} + 4 \hat{j}) \cdot (\hat{i} + \hat{j})}{q (0 \hat{i} + 4 \hat{j}) \cdot (\hat{i} + \hat{j})} = \frac{1}{2}$

$∴$ $V_{new} = \frac{V_{0}}{2}$

Q 2 :

A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. [2007]

A potential difference appears between the two cylinders when a charge density is given to the inner cylinder.
A potential difference appears between the two cylinders when a charge density is given to the outer cylinder.
No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders.
No potential difference appears between the two cylinders when same charge density is given to both the cylinders.

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(1)

According to Gauss's theorem, the electric field between two cylinders is $E = \frac{λ}{2 π ε_{0} r} .$ This electric field will produce a potential difference $d V = - \vec{E} \cdot d \vec{r} .$ When a charge density is given to the inner cylinder, the potential developed at its surface is different from that on the outer cylinder. This is because the potential decreases with distance for a charged conducting cylinder when the point of consideration is outside the cylinder.

But when a charge density is given to the outer cylinder, it will change its potential by the same amount as that of the inner cylinder. Hence, no potential difference will be produced between the cylinders.

Q 3 :

A uniform electric field pointing in positive $x$ -direction exists in a region. Let $A$ be the origin, $B$ be the point on the $x$ -axis at $x = + 1 cm$ and $C$ be the point on the $y$ -axis at $y = + 1 cm$ . Then the potentials at the points $A, B$ and $C$ satisfy: [2001]

$V_{A} < V_{B}$
$V_{A} > V_{B}$
$V_{A} < V_{C}$
$V_{A} > V_{C}$

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(2)

As we move along the direction of the electric field, the potential decreases.

$∴$ $V_{A} > V_{B}$ and $V_{A} = V_{C}$ on equipotential surface.

[IMAGE 705]

Q 4 :

An infinitely long thin wire, having a uniform charge density per unit length of $5 nC/m$ , is passing through a spherical shell of radius 1 m, as shown in the figure. A $10 nC$ charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points $P$ and $R$ , in Volt, is ________.

[Given: In SI units $\frac{1}{4 π ε_{0}} = 9 \times 10^{9}$ , $\ln 2 = 0.7$ . Ignore the area pierced by the wire.] [2024]

[IMAGE 706]

(171)

$Due to wire, d V = - E . d x$

$\int_{V_{P}}^{V_{R}} d V = - \int_{0.5}^{2} \frac{2 k λ}{x} d x$

$V_{R} - V_{P} = - 2 k λ \ln (\frac{2}{0.5})$

$= - 2 \times 9 \times 10^{9} \times 3 \times 10^{- 9} \times 2 \times 0.7 = - 126 V$

$Due to sphere,$

$V_{R} - V_{P} = \frac{k Q}{2} - \frac{k Q}{1} = - \frac{k Q}{2}$

$= \frac{- 9 \times 10^{9} \times 10 \times 10^{- 9}}{2} = - 45 V$

$V_{R} - V_{P} = - 126 - 45 = - 171 V$

$or, V_{P} - V_{R} = 171 V$

Q 5 :

Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch as shown in the figure. Charged oil drops of density $900 {kg m}^{- 3}$ are released through a tiny hole at the center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to apply a voltage of 200 V across the discs. As a result, an oil drop of radius $8 \times 10^{- 7} m$ stops moving vertically and floats between the discs. The number of electrons present in this oil drop is _____. (Neglect the buoyancy force, take acceleration due to gravity $= 10 {ms}^{- 2}$ and charge on an electron $(e) = 1.6 \times 10^{- 19} C$ ) [2020]

[IMAGE 707]

(6)

Let $n$ number of electrons present in the oil drop.

Electric field, $E = \frac{V}{d} = \frac{200}{0.01} = 2 \times 10^{4} V/m$

When terminal velocity is achieved,

$q E = m g$

[IMAGE 708]

As $q = n e$ and $m = \frac{4}{3} π R^{3} ρ,$

therefore, $n (1.6 \times 10^{- 19}) (2 \times 10^{4}) = \frac{4 π}{3} {(8 \times 10^{- 7})}^{3} (900) (10)$

$\Rightarrow n = \frac{4}{3} π \times \frac{{(8 \times 10^{- 7})}^{3} (900) (10)}{(2 \times 10^{4}) (1.6 \times 10^{- 19})}$

$∴$ $n = 6$

Q 6 :

A particle, of mass $10^{- 3} kg$ and charge $1.0 C$ , is initially at rest. At time $t = 0$ , the particle comes under the influence of an electric field $\vec{E} (t) = E_{0} \sin (ω t) \hat{i},$ where $E_{0} = 1.0 {NC}^{- 1}$ and $ω = 10^{3} {rad s}^{- 1}$ . Consider the effect of only the electrical force on the particle. Then the maximum speed, in ${ms}^{- 1}$ , attained by the particle at subsequent times is _______. [2018]

(2)

$∵$ $F = m a$ $∴$ $q E = m \frac{d v}{d t}$

$\Rightarrow d v = \frac{q E d t}{m} = \frac{q \sin (1000 t) \hat{i}}{m} d t$

$(∵ E = \sin (10^{3} t) \hat{i} is given)$

$∴$ $\int_{0}^{v} d v = \frac{q}{m} \int_{0}^{π / ω} \sin (1000 t) d t$ $[Maximum speed is attained at \frac{T}{2} = \frac{2 π}{ω \times 2}]$

$∴$ $v = - \frac{q}{m} {[\frac{\cos (1000 t)}{1000}]}_{0}^{π / ω}$ $= - \frac{1}{10^{- 3}} {[\frac{\cos (1000 t)}{1000}]}_{0}^{π / ω}$

$(∵ m = 10^{- 3} kg, q = 1 C, E_{0} = 1 {N C}^{- 1}, ω = 10^{3} {rad s}^{- 1})$ given

$∴$ $v = - [\cos (1000 \times \frac{π}{1000}) - \cos 0]$ $= - [- 1 - 1] = 2 {ms}^{- 1}$

Hence, the maximum speed attained by the particle.

Q 7 :

Two point charges $- Q$ and $+ \frac{Q}{\sqrt{3}}$ are placed in the $x y$ -plane at the origin $(0, 0)$ and a point $(2, 0)$ , respectively, as shown in the figure. This results in an equipotential circle of radius $R$ and potential $V = 0$ in the $x y$ -plane with its center at $(b, 0)$ . All lengths are measured in meters.

[IMAGE 709]

Q. The value of R is ______ meter. [2021]

(1.73)

Let us consider a point $P$ on the circle.

[IMAGE 710]

$V_{P} = 0 = \frac{k (- Q)}{r_{1}} + \frac{k Q / \sqrt{3}}{r_{2}}$ $\Rightarrow \frac{k Q}{r_{1}} = \frac{k Q / \sqrt{3}}{r_{2}}$

$\Rightarrow \frac{1}{\sqrt{x^{2} + y^{2}}} - \frac{1}{\sqrt{3} \sqrt{{(x - 2)}^{2} + y^{2}}}$

$\Rightarrow 3 {(x - 2)}^{2} + 3 y^{2} = x^{2} + y^{2}$

$\Rightarrow 3 (x^{2} + 4 - 4 x) - x^{2} + 2 y^{2} = 0$

$\Rightarrow 2 x^{2} + 12 - 12 x + 2 y^{2} = 0$

$\Rightarrow x^{2} + 6 - 6 x + y^{2} = 0$

$\Rightarrow {(x - 3)}^{2} + y^{2} = {(\sqrt{3})}^{2}$

or ${(x - b)}^{2} + y^{2} = {(\sqrt{3})}^{2} = R^{2}$

$∴$ $R = \sqrt{3} \approx 1.73$ and $b = 3$

Q 8 :

Two point charges $- Q$ and $+ \frac{Q}{\sqrt{3}}$ are placed in the $x y$ -plane at the origin $(0, 0)$ and a point $(2, 0)$ , respectively, as shown in the figure. This results in an equipotential circle of radius $R$ and potential $V = 0$ in the $x y$ -plane with its center at $(b, 0)$ . All lengths are measured in meters.

[IMAGE 711]

Q. The value of $b$ is _________ meter. [2021]

(3)

Let us consider a point $P$ on the circle.

[IMAGE 712]

$V_{P} = 0 = \frac{k (- Q)}{r_{1}} + \frac{k Q / \sqrt{3}}{r_{2}}$ $\Rightarrow \frac{k Q}{r_{1}} = \frac{k Q / \sqrt{3}}{r_{2}}$

$\Rightarrow \frac{1}{\sqrt{x^{2} + y^{2}}} - \frac{1}{\sqrt{3} \sqrt{{(x - 2)}^{2} + y^{2}}}$

$\Rightarrow 3 {(x - 2)}^{2} + 3 y^{2} = x^{2} + y^{2}$

$\Rightarrow 3 (x^{2} + 4 - 4 x) - x^{2} + 2 y^{2} = 0$

$\Rightarrow 2 x^{2} + 12 - 12 x + 2 y^{2} = 0$

$\Rightarrow x^{2} + 6 - 6 x + y^{2} = 0$

$\Rightarrow {(x - 3)}^{2} + y^{2} = {(\sqrt{3})}^{2}$

or ${(x - b)}^{2} + y^{2} = {(\sqrt{3})}^{2} = R^{2}$

$∴$ $R = \sqrt{3} \approx 1.73$ and $b = 3$

Q 9 :

Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure, the sheets carry uniform surface charge densities which are indicated in terms of $σ_{0}$ . The separation between any two consecutive sheets is $1 μ m$ . The various regions between the sheets are denoted as 1, 2, 3, 4 and 5. If $σ_{0} = 9 μ {C/m}^{2}$ , then which of the following statements is/are correct:

(Take permittivity of free space $ε_{0} = 9 \times 10^{- 12}$ F/m) [2025]

[IMAGE 713]

In region 4 of the configuration I, the magnitude of the electric field is zero.
In region 3 of the configuration II, the magnitude of the electric field is $\frac{σ_{0}}{ε_{0}} .$
Potential difference between the first and the last sheets of the configuration I is 5 V.
Potential difference between the first and the last sheets of the configuration II is zero.

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(1)

In region 4, configuration I

${(E_{4})}_{I} = \frac{σ_{0}}{2 ε_{0}} [1 - 1 + 1 - 1 - 1 + 1] = 0$

[IMAGE 714]

${(V_{First})}_{I} = \frac{- σ_{0}}{2 ε_{0}} [- 1 + 2 - 3 + 4 - 5] d$

$= \frac{- σ_{0}}{2 ε_{0}} (- 3) d = \frac{3 σ_{0} d}{2 ε_{0}}$

${(V_{Last})}_{I} = \frac{- σ_{0}}{2 ε_{0}} [1 - 2 + 3 - 4 + 5] d$

$= \frac{σ_{0}}{2 ε_{0}} (- 3) d = - \frac{3 σ_{0} d}{2 ε_{0}}$

${(V_{First} - V_{Last})}_{I} = \frac{3 σ_{0} d}{ε_{0}}$

$= \frac{3 \times 9 \times 10^{- 6} \times 1 \times 10^{- 6}}{9 \times 10^{- 12}} = 3 V$

In region 3, configuration II

${(E_{3})}_{I I} = \frac{σ_{0}}{2 ε_{0}} [\frac{1}{2} - 1 + 1 + 1 - 1 + \frac{1}{2}] = \frac{σ_{0}}{2 ε_{0}}$

${(V_{First})}_{I I} = \frac{- σ_{0}}{2 ε_{0}} [- 1 + 2 - 3 + 4 - \frac{5}{2}] d$

[IMAGE 715]

$= \frac{- σ_{0}}{2 ε_{0}} [2 - 2.5] d = \frac{σ_{0} d}{4 ε_{0}}$

${(V_{Last})}_{I I} = \frac{- σ_{0}}{2 ε_{0}} [1 - 2 + 3 - 4 + \frac{5}{2}] d$

$= \frac{- σ_{0}}{2 ε_{0}} [6.5 - 6] d = \frac{- σ_{0} d}{4 ε_{0}}$

${(V_{First} - V_{Last})}_{I I} = \frac{σ_{0} d}{2 ε_{0}} \neq 0$

Q 10 :

Six charges are placed around a regular hexagon of side length $a$ as shown in the figure. Five of them have charge $q$ , and the remaining one has charge $x$ . The perpendicular from each charge to the nearest hexagon side passes through the center $O$ of the hexagon and is bisected by the side.

Which of the following statement(s) is(are) correct in SI units? [2022]

[IMAGE 716]

When $x = q$ , the magnitude of the electric field at $O$ is zero.
When $x = - q$ , the magnitude of the electric field at $O$ is $\frac{q}{6 π ε_{0} a^{2}} .$
When $x = 2 q$ , the potential at $O$ is $\frac{7 q}{4 \sqrt{3} π ε_{0} a} .$
When $x = - 3 q$ , the potential at $O$ is $- \frac{3 q}{4 \sqrt{3} π ε_{0} a} .$

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Select one or more options

(1, 2, 3)

[IMAGE 717]

(1) When $x = q$ , the situation is symmetric.

So, electric field at $O$ is zero.

$\Rightarrow (1) is correct.$

(2) When $x = - q$ , then $E_{0} = \frac{1}{4 π ε_{0}} \frac{q}{{(\sqrt{3} a)}^{2}} \times 2$

$\Rightarrow E_{0} = \frac{q}{6 π ε_{0} a^{2}}$

$\Rightarrow (2) is correct.$

(3) When $x = 2 q$ ,

$V_{0} = 5 \times \frac{K q}{\sqrt{3} a} + \frac{K (2 q)}{\sqrt{3} a}$ $\Rightarrow V_{0} = \frac{7 q}{4 \sqrt{3} π ε_{0} a}$

$\Rightarrow (3) is correct.$

(4) When $x = - 3 q$ ,

$V_{0} = 5 \times \frac{K q}{\sqrt{3} a} + \frac{K (- 3 q)}{\sqrt{3} a}$ $= \frac{1}{4 π ε_{0}} \frac{2 q}{\sqrt{3} a} = \frac{q}{2 \sqrt{3} π ε_{0} a}$

$\Rightarrow (4) is incorrect.$

Q 11 :

In the figure, the inner (shaded) region $A$ represents a sphere of radius $r_{A} = 1$ , within which the electrostatic charge density varies with the radial distance $r$ from the center as $ρ_{A} = k r,$ where $k$ is positive.

In the spherical shell $B$ of outer radius $r_{B}$ , the electrostatic charge density varies as $ρ_{B} = \frac{2 k}{r} .$

Assume that dimensions are taken care of. All physical quantities are in their SI units.

Which of the following statement(s) is (are) correct? [2022]

[IMAGE 718]

If $r_{B} = \sqrt{\frac{3}{2}}$ , then the electric field is zero everywhere outside $B$ .
If $r_{B} = \frac{3}{2}$ , then the electric potential just outside $B$ is $\frac{k}{ε_{0}} .$
If $r_{B} = 2$ , then the total charge of the configuration is $15 π k .$
If $r_{B} = \frac{5}{2}$ , then the magnitude of the electric field just outside $B$ is $\frac{13 π k}{ε_{0}} .$

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(2)

We have

[IMAGE 719]

$q_{A} = \int_{0}^{1} k r (4 π r^{2} d r) = 4 π k \int_{0}^{1} r^{3} d r = \frac{4 π k}{4} {[r^{4}]}_{0}^{1} = π k$

$q_{B} = \int_{1}^{r} \frac{2 k}{r} 4 π r^{2} d r = 8 π k \int_{1}^{r} r d r = 8 π k (\frac{r^{2} - 1}{2})$

$= 4 π k r^{2} - 4 π k$

$∴$ $q_{net} = q_{A} + q_{B} = 4 π k r^{2} - 3 π k = π k (4 r^{2} - 3)$

(1) $E_{net} = 0 \Rightarrow q_{net} = 0 \Rightarrow 4 r^{2} - 3 = 0$ $\Rightarrow r = \frac{\sqrt{3}}{2}$

So (1) is incorrect.

(2) $V = \frac{1}{4 π ε_{0}} \frac{q_{net}}{r} = \frac{1}{4 π ε_{0}} \frac{π k (4 r^{2} - 3)}{r}$

$\Rightarrow V = \frac{k}{4 ε_{0}} (4 r - \frac{3}{r})$ $\Rightarrow V = \frac{k}{4 ε_{0}} (4 \times \frac{3}{2} - \frac{3}{3} \times 2)$

$= \frac{k}{4 ε_{0}} (6 - 2) = \frac{k}{ε_{0}}$

So (2) is correct.

(3) If $r_{B} = 2,$ i.e. $r = 2$

then, $q_{net} = π k (16 - 3) = 13 π k .$

So (3) is incorrect.

(4) $E = \frac{k q_{net}}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{π k (4 r^{2} - 3)}{r^{2}}$ $= \frac{k}{4 ε_{0}} (4 - \frac{3}{r^{2}})$

$E = \frac{k}{4 ε_{0}} (4 - \frac{3}{25} \times 4)$ $= \frac{k}{4 ε_{0}} [\frac{88}{25}] = \frac{22 k}{25 ε_{0}}$

Hence (4) is correct.

Q 12 :

A disk of radius $R$ with uniform positive charge density $σ$ is placed on the $x y$ -plane with its center at the origin.

The Coulomb potential along the $z$ -axis is $V (z) = \frac{σ}{2 ε_{0}} (\sqrt{R^{2} + z^{2} - z}) .$ A particle of positive charge $q$ is placed initially at rest at a point on the $z$ -axis with $z = z_{0}$ and $z_{0} > 0$ . In addition to the Coulomb force, the particle experiences a vertical force $\vec{F} = - c \hat{k}$ with $c > 0$ . Let $β = \frac{2 c ε_{0}}{q σ} .$ Which of the following statement(s) is(are) correct? [2022]

For $β = \frac{1}{4}$ and $z_{0} = \frac{25}{7} R,$ the particle reaches the origin.
For $β = \frac{1}{4}$ and $z_{0} = \frac{3}{7} R,$ the particle reaches the origin.
For $β = \frac{1}{4}$ and $z_{0} = \frac{R}{\sqrt{3}},$ the particle returns back to $z = z_{0} .$
For $β > 1$ and $z_{0} > 0$ , the particle always reaches the origin.

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Select one or more options

(1, 3, 4)

[IMAGE 720]

Particle will reach the origin only if

$Δ K \geq 0$

$\Rightarrow W_{el} + W_{ext} > 0$

$\Rightarrow \frac{σ q}{2 ε_{0}} [(\sqrt{R^{2} + 0^{2}} - 0) + (\sqrt{R^{2} + Z_{0}^{2}} - Z_{0})] + C Z_{0} \geq 0$

$\Rightarrow \frac{σ q}{2 ε_{0}} (R - \sqrt{R^{2} + Z_{0}^{2}} + Z_{0}) + C Z_{0} \geq 0$

$\Rightarrow \frac{σ q}{2 ε_{0}} C (R - \sqrt{R^{2} + Z_{0}^{2}} + Z_{0}) + \frac{C Z_{0}}{C} \geq 0$

$\Rightarrow \frac{1}{β} (R + Z_{0} - \sqrt{R^{2} + Z_{0}^{2}}) + Z_{0} \geq 0$

Substitute $Z_{0}$ and $β$ , and check the condition.

If $Δ K > 0$ , the particle will reach the origin; otherwise, it will not reach the origin.

Q 13 :

Two non-conducting spheres of radii $R_{1}$ and $R_{2}$ and carrying uniform volume charge densities $+ ρ$ and $- ρ$ , respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region [2013]

[IMAGE 721]

The electrostatic field is zero
The electrostatic potential is constant
The electrostatic field is constant in magnitude
The electrostatic field has the same direction

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Select one or more options

(3, 4)

Electrostatic field at P is

[IMAGE 722]

${\vec{E}}_{P} = {\vec{E}}_{1} + {\vec{E}}_{2} = \frac{ρ}{3 ε_{0}} \vec{C_{1} P} + \frac{(- ρ)}{3 ε_{0}} \vec{C_{2} P}$

$= \frac{ρ}{3 ε_{0}} (\vec{C_{1} P} - \vec{C_{2} P})$ $= \frac{ρ}{3 ε_{0}} (\vec{C_{1} P} + \vec{P C_{2}})$ $= \frac{ρ}{3 ε_{0}} (\vec{C_{1} C_{2}})$

Q 14 :

Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$ , as shown in the figure. Given that $K = \frac{1}{4 π ε_{0}} \frac{q}{L^{2}},$ which of the following statement(s) is (are) correct? [2012]

[IMAGE 723]

The electric field at O is 6K along OD
The potential at O is zero
The potential at all points on the line PR is same
The potential at all points on the line ST is same

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Select one or more options

(1, 2, 3)

Here $\frac{| {\vec{E}}_{A} |}{2} = | {\vec{E}}_{B} | = | {\vec{E}}_{C} | = \frac{| {\vec{E}}_{D} |}{2} = | {\vec{E}}_{E} | = | {\vec{E}}_{F} | = K$

$∴$ $E_{O} = E_{A} + E_{D} + (E_{F} + E_{C}) \cos 60 ° + (E_{B} + E_{C}) \cos 60 °$

$= 2 K + 2 K + (K + K) (\frac{1}{2}) + (K + K) (\frac{1}{2})$

$= 2 K + 2 K + K + K = 6 K$

[IMAGE 724]

Electric potential at center, O

$V_{O} = \frac{1}{4 π ε_{0} L} [2 q + q + q - q - q - 2 q] = 0$

Potential at all points on the line PR is the same, but not on line ST.

PR is the perpendicular bisector (the equatorial line) for the electric dipoles AB, FE, and BC. Therefore, the electric potential will be zero at any point on PR.

Q 15 :

Which of the following statement(s) is/are correct? [2011]

If the electric field due to a point charge varies as $r^{- 2.5}$ instead of $r^{- 2}$ , then the Gauss law will still be valid.
The Gauss law can be used to calculate the field distribution around an electric dipole.
If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same.
The work done by the external force in moving a unit positive charge from point $A$ at potential $V_{A}$ to point $B$ at potential $V_{B}$ is $(V_{B} - V_{A})$ .

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Select one or more options

(3, 4)

(1) Gauss's law is valid only when $E \propto r^{- 2} .$

(2) Gauss's law cannot be used to calculate the field distribution around an electric dipole.

(3) (3) is correct, as between two point charges we will get a point where the electric field due to the two point charges cancels out each other. If two point charges are of opposite sign, then the two fields are along the same direction; hence they cannot be zero.

(4) $W_{A B} = q (Δ V) = 1 (V_{B} - V_{A})$ or, $W_{A B} = V_{B} - V_{A} .$

Q 16 :

A spherical metal shell A of radius $R_{A}$ and a solid metal sphere B of radius $R_{B}$ $(R_{B} < R_{A})$ are kept far apart and each is given charge $' + Q'$ . Now they are connected by a thin metal wire. Then [2011]

$E_{A}^{inside} = 0$
$Q_{A} > Q_{B}$
$\frac{σ_{A}}{σ_{B}} = \frac{R_{B}}{R_{A}}$
$E_{A}^{on surface} < E_{B}^{on surface}$

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Select one or more options

(1, 2, 3, 4)

(1) Electric field inside a spherical metallic shell with charge on the surface is always zero, i.e., $E_{A}^{inside} = 0 .$

(2) When the shells are connected with a thin metal wire, then electric potentials will be equal, i.e., $V_{A} = V_{B} = V .$

$∴$ $\frac{1}{4 π ε_{0}} \frac{Q_{A}}{R_{A}} = \frac{1}{4 π ε_{0}} \frac{Q_{B}}{R_{B}} = V$ $∵$ $R_{A} > R_{B}$ $∴$ $Q_{A} > Q_{B}$

(3) As $\frac{σ_{A}}{σ_{B}} = \frac{\frac{Q_{A}}{4 π R_{A}^{2}}}{\frac{Q_{B}}{4 π R_{B}^{2}}} = \frac{R_{B}^{2}}{R_{A}^{2}} \cdot \frac{Q_{A}}{Q_{B}}$ $= \frac{R_{B}^{2}}{R_{A}^{2}} \cdot \frac{4 π ε_{0} R_{A} V}{4 π ε_{0} R_{B} V} = \frac{R_{B}}{R_{A}}$

(4) $E_{A} = \frac{σ_{A}}{ε_{0}}$ and $E_{B} = \frac{σ_{B}}{ε_{0}}$

$\frac{E_{A}}{E_{B}} = \frac{σ_{A}}{σ_{B}} = \frac{R_{B}}{R_{A}} < 1$

$∴$ $E_{A} < E_{B}$

Q 17 :

A spherically symmetric charge system is centered at the origin. Given electric potential

$V = \frac{Q}{4 π ε_{0} R_{0}} (r \leq R_{0}),$ $V = \frac{Q}{4 π ε_{0} r} (r > R_{0}) .$ [2006]

[IMAGE 725]

Within $r = 2 R_{0}$ , total enclosed net charge is $Q$
Electric field is discontinuous at $r = R_{0}$
Charge is only present at $r = R_{0}$
Electrostatic energy is zero for $r < R_{0}$

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Select one or more options

(1, 2, 4)

[IMAGE 726]

The given $V - r$ graph is of a charged conducting sphere of radius $R_{0}$

(1) The whole charge $Q$ will be enclosed in a sphere of diameter $2 R_{0}$ .

(2) Electric field $E = 0$ inside the sphere. Hence electric field is discontinued at $r = R_{0}$ .

(3) Changes in $V$ and $E$ are continuously present for $r > R_{0}$ .

Option (3) is incorrect.

(4) For $r < R_{0}$ , the potential $V$ is constant and the electric intensity is zero. Obviously the electrostatic energy is zero for $r < R_{0}$

Q 18 :

STATEMENT-1 : For practical purposes, the earth is used as a reference at zero potential in electrical circuits.

STATEMENT-2 : The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by $\frac{Q}{4 π ε_{0} R} .$ [2008]

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
Statement-1 is True, Statement-2 is False
Statement-1 is False, Statement-2 is True

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The earth is used as a reference at zero potential in electrical circuits for practical purposes.

For a spherical capacitor, capacitance $C = 4 π ε_{0} R$ and electric potential $V = \frac{Q}{C} = \frac{Q}{4 π ε_{0} R} .$

Topic Question Set

Q 2 :

A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. [2007]

Q 3 :

A uniform electric field pointing in positive $x$ -direction exists in a region. Let $A$ be the origin, $B$ be the point on the $x$ -axis at $x = + 1 cm$ and $C$ be the point on the $y$ -axis at $y = + 1 cm$ . Then the potentials at the points $A, B$ and $C$ satisfy: [2001]

Q 13 :

Two non-conducting spheres of radii $R_{1}$ and $R_{2}$ and carrying uniform volume charge densities $+ ρ$ and $- ρ$ , respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region [2013]

[IMAGE 721]

Q 14 :

Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$ , as shown in the figure. Given that $K = \frac{1}{4 π ε_{0}} \frac{q}{L^{2}},$ which of the following statement(s) is (are) correct? [2012]

[IMAGE 723]

Q 15 :

Which of the following statement(s) is/are correct? [2011]

Q 16 :

A spherical metal shell A of radius $R_{A}$ and a solid metal sphere B of radius $R_{B}$ $(R_{B} < R_{A})$ are kept far apart and each is given charge $' + Q'$ . Now they are connected by a thin metal wire. Then [2011]

Q 17 :

A spherically symmetric charge system is centered at the origin. Given electric potential

$V = \frac{Q}{4 π ε_{0} R_{0}} (r \leq R_{0}),$ $V = \frac{Q}{4 π ε_{0} r} (r > R_{0}) .$ [2006]

[IMAGE 725]

Q 18 :

STATEMENT-1 : For practical purposes, the earth is used as a reference at zero potential in electrical circuits.

STATEMENT-2 : The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by $\frac{Q}{4 π ε_{0} R} .$ [2008]

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Topic Question Set

Q 2 : A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. [2007]

Q 3 : A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x=+1 cm and C be the point on the y-axis at y=+1 cm. Then the potentials at the points A, B and C satisfy: [2001]

Q 13 : Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge densities +ρ and −ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region [2013] [IMAGE 721]

Q 14 : Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the figure. Given that K=14πε0qL2, which of the following statement(s) is (are) correct? [2012] [IMAGE 723]

Q 15 : Which of the following statement(s) is/are correct? [2011]

Q 16 : A spherical metal shell A of radius RA and a solid metal sphere B of radius RB(RB<RA) are kept far apart and each is given charge '+Q'. Now they are connected by a thin metal wire. Then [2011]

Q 17 : A spherically symmetric charge system is centered at the origin. Given electric potential V=Q4πε0R0(r≤R0), V=Q4πε0r(r>R0). [2006] [IMAGE 725]

Q 18 : STATEMENT-1 : For practical purposes, the earth is used as a reference at zero potential in electrical circuits. STATEMENT-2 : The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by Q4πε0R. [2008]

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Q 2 :

A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. [2007]

Q 3 :

A uniform electric field pointing in positive $x$ -direction exists in a region. Let $A$ be the origin, $B$ be the point on the $x$ -axis at $x = + 1 cm$ and $C$ be the point on the $y$ -axis at $y = + 1 cm$ . Then the potentials at the points $A, B$ and $C$ satisfy: [2001]

Q 13 :

Two non-conducting spheres of radii $R_{1}$ and $R_{2}$ and carrying uniform volume charge densities $+ ρ$ and $- ρ$ , respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region [2013]

[IMAGE 721]

Q 14 :

Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$ , as shown in the figure. Given that $K = \frac{1}{4 π ε_{0}} \frac{q}{L^{2}},$ which of the following statement(s) is (are) correct? [2012]

[IMAGE 723]

Q 15 :

Which of the following statement(s) is/are correct? [2011]

Q 16 :

A spherical metal shell A of radius $R_{A}$ and a solid metal sphere B of radius $R_{B}$ $(R_{B} < R_{A})$ are kept far apart and each is given charge $' + Q'$ . Now they are connected by a thin metal wire. Then [2011]

Q 17 :

A spherically symmetric charge system is centered at the origin. Given electric potential

$V = \frac{Q}{4 π ε_{0} R_{0}} (r \leq R_{0}),$ $V = \frac{Q}{4 π ε_{0} r} (r > R_{0}) .$ [2006]

[IMAGE 725]

Q 18 :

STATEMENT-1 : For practical purposes, the earth is used as a reference at zero potential in electrical circuits.

STATEMENT-2 : The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by $\frac{Q}{4 π ε_{0} R} .$ [2008]