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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359413 A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in $V$ ) shown in the figure is:
$($ Take $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} S I$ units $)$
supporting img

1

3 \times 10^{5}

2

1 \times 10^{5}

3

0.5 \times 10^{5}

4 Zero

Explanation:

For uniformly charged spherical shell,
$V = \frac{k q}{R} ($ For $r \leq R)$
$∴ V_{C} = V_{P}$
$V_{C} - V_{P} =$ Zero

NEET - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359414 Statement A :
Two adjacent conductors of unequal dimensions, carrying the same positive charge have non-zero potential difference between them.
Statement B :
The potential of a conductor depends upon only the charge given to it.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

Let us consider two spherical shells of radii $r_{1}$ and $r_{2}$ which are possessing the same positive charge $Q$ . Therefore the potential on the surface of each conductor will be,
$V_{1} = \frac{k Q}{r_{1}}$ and $V_{2} = \frac{k Q}{r_{2}}$ since, $r_{1} \neq r_{2}$
$V_{1} \neq V_{2}$ are consequently there will be a potential difference between the two conductors.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359415 Assertion :
For practical purposes, the earth is used as a reference at zero potential in electrical circuits.
Reason :
The electrical potential of a sphere of radius $R$ with charge $Q$ uniformly distributed on the surface is given by $\frac{Q}{4 π ε_{0} R}$ . As $R$ is very large and $Q$ is very small (total charge of earth) so the net potential of earth is approximately zero.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359416 The potential of a large liquid drop when eight liquid drops are combined is $20 V$ . Then the potential of each single drop was

1

10 V

2

7.5 V

3

5 V

4

2.5 V

Explanation:

Let $R$ be radius of the big drop and $r$ be radius of each small drop.
$As volume of the big drop = volume of$
$eight small drops$
$∴ \frac{4}{3} π R^{3} = 8 (\frac{4}{3} π r^{3})$
$R = 2 r (1)$
If $q$ be charge on each small drop, then charge on the big drop is
$Q = 8 q (2)$
The potential of each small drop is $V = \frac{1}{4 π ε_{0}} \frac{q}{r} (3)$
and that of the big drop is
$V^{'} = \frac{1}{4 π ε_{0}} \frac{Q}{R} (4)$
Dividing eqn. (3) by eqn. (4), we get
$\frac{V}{V^{'}} = \frac{\frac{1}{4 π ε_{0}} \frac{q}{r}}{\frac{1}{4 π ε_{0}} \frac{Q}{R}} = \frac{q}{Q} \frac{R}{r}$
$= \frac{q}{8 q} \frac{2 r}{r} = \frac{1}{4} (Using eqns. (1) and (2))$
$V = \frac{1}{4} V^{'}$
As $V^{'} = 20 V$ (given)
$∴ V = \frac{1}{4} (20 V) = 5 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359417 A hollow conducting sphere is placed in an electric field produced by a point charge placed at $P$ as shown in figure. Let $V_{A}, V_{B}, V_{C}$ be the potential at points $A$ , $B$ and $C$ respectively. Then
supporting img

1

V_{B} > V_{C}

2

V_{A} > V_{B}

3

V_{C} > V_{B}

4

V_{A} = V_{C}

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359413 A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in $V$ ) shown in the figure is:
$($ Take $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} S I$ units $)$
supporting img

1

3 \times 10^{5}

2

1 \times 10^{5}

3

0.5 \times 10^{5}

4 Zero

Explanation:

For uniformly charged spherical shell,
$V = \frac{k q}{R} ($ For $r \leq R)$
$∴ V_{C} = V_{P}$
$V_{C} - V_{P} =$ Zero

NEET - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359414 Statement A :
Two adjacent conductors of unequal dimensions, carrying the same positive charge have non-zero potential difference between them.
Statement B :
The potential of a conductor depends upon only the charge given to it.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

Let us consider two spherical shells of radii $r_{1}$ and $r_{2}$ which are possessing the same positive charge $Q$ . Therefore the potential on the surface of each conductor will be,
$V_{1} = \frac{k Q}{r_{1}}$ and $V_{2} = \frac{k Q}{r_{2}}$ since, $r_{1} \neq r_{2}$
$V_{1} \neq V_{2}$ are consequently there will be a potential difference between the two conductors.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359415 Assertion :
For practical purposes, the earth is used as a reference at zero potential in electrical circuits.
Reason :
The electrical potential of a sphere of radius $R$ with charge $Q$ uniformly distributed on the surface is given by $\frac{Q}{4 π ε_{0} R}$ . As $R$ is very large and $Q$ is very small (total charge of earth) so the net potential of earth is approximately zero.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359416 The potential of a large liquid drop when eight liquid drops are combined is $20 V$ . Then the potential of each single drop was

1

10 V

2

7.5 V

3

5 V

4

2.5 V

Explanation:

Let $R$ be radius of the big drop and $r$ be radius of each small drop.
$As volume of the big drop = volume of$
$eight small drops$
$∴ \frac{4}{3} π R^{3} = 8 (\frac{4}{3} π r^{3})$
$R = 2 r (1)$
If $q$ be charge on each small drop, then charge on the big drop is
$Q = 8 q (2)$
The potential of each small drop is $V = \frac{1}{4 π ε_{0}} \frac{q}{r} (3)$
and that of the big drop is
$V^{'} = \frac{1}{4 π ε_{0}} \frac{Q}{R} (4)$
Dividing eqn. (3) by eqn. (4), we get
$\frac{V}{V^{'}} = \frac{\frac{1}{4 π ε_{0}} \frac{q}{r}}{\frac{1}{4 π ε_{0}} \frac{Q}{R}} = \frac{q}{Q} \frac{R}{r}$
$= \frac{q}{8 q} \frac{2 r}{r} = \frac{1}{4} (Using eqns. (1) and (2))$
$V = \frac{1}{4} V^{'}$
As $V^{'} = 20 V$ (given)
$∴ V = \frac{1}{4} (20 V) = 5 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359417 A hollow conducting sphere is placed in an electric field produced by a point charge placed at $P$ as shown in figure. Let $V_{A}, V_{B}, V_{C}$ be the potential at points $A$ , $B$ and $C$ respectively. Then
supporting img

1

V_{B} > V_{C}

2

V_{A} > V_{B}

3

V_{C} > V_{B}

4

V_{A} = V_{C}

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359413 A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in $V$ ) shown in the figure is:
$($ Take $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} S I$ units $)$
supporting img

1

3 \times 10^{5}

2

1 \times 10^{5}

3

0.5 \times 10^{5}

4 Zero

Explanation:

For uniformly charged spherical shell,
$V = \frac{k q}{R} ($ For $r \leq R)$
$∴ V_{C} = V_{P}$
$V_{C} - V_{P} =$ Zero

NEET - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359414 Statement A :
Two adjacent conductors of unequal dimensions, carrying the same positive charge have non-zero potential difference between them.
Statement B :
The potential of a conductor depends upon only the charge given to it.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

Let us consider two spherical shells of radii $r_{1}$ and $r_{2}$ which are possessing the same positive charge $Q$ . Therefore the potential on the surface of each conductor will be,
$V_{1} = \frac{k Q}{r_{1}}$ and $V_{2} = \frac{k Q}{r_{2}}$ since, $r_{1} \neq r_{2}$
$V_{1} \neq V_{2}$ are consequently there will be a potential difference between the two conductors.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359415 Assertion :
For practical purposes, the earth is used as a reference at zero potential in electrical circuits.
Reason :
The electrical potential of a sphere of radius $R$ with charge $Q$ uniformly distributed on the surface is given by $\frac{Q}{4 π ε_{0} R}$ . As $R$ is very large and $Q$ is very small (total charge of earth) so the net potential of earth is approximately zero.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359416 The potential of a large liquid drop when eight liquid drops are combined is $20 V$ . Then the potential of each single drop was

1

10 V

2

7.5 V

3

5 V

4

2.5 V

Explanation:

Let $R$ be radius of the big drop and $r$ be radius of each small drop.
$As volume of the big drop = volume of$
$eight small drops$
$∴ \frac{4}{3} π R^{3} = 8 (\frac{4}{3} π r^{3})$
$R = 2 r (1)$
If $q$ be charge on each small drop, then charge on the big drop is
$Q = 8 q (2)$
The potential of each small drop is $V = \frac{1}{4 π ε_{0}} \frac{q}{r} (3)$
and that of the big drop is
$V^{'} = \frac{1}{4 π ε_{0}} \frac{Q}{R} (4)$
Dividing eqn. (3) by eqn. (4), we get
$\frac{V}{V^{'}} = \frac{\frac{1}{4 π ε_{0}} \frac{q}{r}}{\frac{1}{4 π ε_{0}} \frac{Q}{R}} = \frac{q}{Q} \frac{R}{r}$
$= \frac{q}{8 q} \frac{2 r}{r} = \frac{1}{4} (Using eqns. (1) and (2))$
$V = \frac{1}{4} V^{'}$
As $V^{'} = 20 V$ (given)
$∴ V = \frac{1}{4} (20 V) = 5 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359417 A hollow conducting sphere is placed in an electric field produced by a point charge placed at $P$ as shown in figure. Let $V_{A}, V_{B}, V_{C}$ be the potential at points $A$ , $B$ and $C$ respectively. Then
supporting img

1

V_{B} > V_{C}

2

V_{A} > V_{B}

3

V_{C} > V_{B}

4

V_{A} = V_{C}

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359413 A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in $V$ ) shown in the figure is:
$($ Take $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} S I$ units $)$
supporting img

1

3 \times 10^{5}

2

1 \times 10^{5}

3

0.5 \times 10^{5}

4 Zero

Explanation:

For uniformly charged spherical shell,
$V = \frac{k q}{R} ($ For $r \leq R)$
$∴ V_{C} = V_{P}$
$V_{C} - V_{P} =$ Zero

NEET - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359414 Statement A :
Two adjacent conductors of unequal dimensions, carrying the same positive charge have non-zero potential difference between them.
Statement B :
The potential of a conductor depends upon only the charge given to it.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

Let us consider two spherical shells of radii $r_{1}$ and $r_{2}$ which are possessing the same positive charge $Q$ . Therefore the potential on the surface of each conductor will be,
$V_{1} = \frac{k Q}{r_{1}}$ and $V_{2} = \frac{k Q}{r_{2}}$ since, $r_{1} \neq r_{2}$
$V_{1} \neq V_{2}$ are consequently there will be a potential difference between the two conductors.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359415 Assertion :
For practical purposes, the earth is used as a reference at zero potential in electrical circuits.
Reason :
The electrical potential of a sphere of radius $R$ with charge $Q$ uniformly distributed on the surface is given by $\frac{Q}{4 π ε_{0} R}$ . As $R$ is very large and $Q$ is very small (total charge of earth) so the net potential of earth is approximately zero.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359416 The potential of a large liquid drop when eight liquid drops are combined is $20 V$ . Then the potential of each single drop was

1

10 V

2

7.5 V

3

5 V

4

2.5 V

Explanation:

Let $R$ be radius of the big drop and $r$ be radius of each small drop.
$As volume of the big drop = volume of$
$eight small drops$
$∴ \frac{4}{3} π R^{3} = 8 (\frac{4}{3} π r^{3})$
$R = 2 r (1)$
If $q$ be charge on each small drop, then charge on the big drop is
$Q = 8 q (2)$
The potential of each small drop is $V = \frac{1}{4 π ε_{0}} \frac{q}{r} (3)$
and that of the big drop is
$V^{'} = \frac{1}{4 π ε_{0}} \frac{Q}{R} (4)$
Dividing eqn. (3) by eqn. (4), we get
$\frac{V}{V^{'}} = \frac{\frac{1}{4 π ε_{0}} \frac{q}{r}}{\frac{1}{4 π ε_{0}} \frac{Q}{R}} = \frac{q}{Q} \frac{R}{r}$
$= \frac{q}{8 q} \frac{2 r}{r} = \frac{1}{4} (Using eqns. (1) and (2))$
$V = \frac{1}{4} V^{'}$
As $V^{'} = 20 V$ (given)
$∴ V = \frac{1}{4} (20 V) = 5 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359417 A hollow conducting sphere is placed in an electric field produced by a point charge placed at $P$ as shown in figure. Let $V_{A}, V_{B}, V_{C}$ be the potential at points $A$ , $B$ and $C$ respectively. Then
supporting img

1

V_{B} > V_{C}

2

V_{A} > V_{B}

3

V_{C} > V_{B}

4

V_{A} = V_{C}

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359413 A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in $V$ ) shown in the figure is:
$($ Take $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} S I$ units $)$
supporting img

1

3 \times 10^{5}

2

1 \times 10^{5}

3

0.5 \times 10^{5}

4 Zero

Explanation:

For uniformly charged spherical shell,
$V = \frac{k q}{R} ($ For $r \leq R)$
$∴ V_{C} = V_{P}$
$V_{C} - V_{P} =$ Zero

NEET - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359414 Statement A :
Two adjacent conductors of unequal dimensions, carrying the same positive charge have non-zero potential difference between them.
Statement B :
The potential of a conductor depends upon only the charge given to it.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

Let us consider two spherical shells of radii $r_{1}$ and $r_{2}$ which are possessing the same positive charge $Q$ . Therefore the potential on the surface of each conductor will be,
$V_{1} = \frac{k Q}{r_{1}}$ and $V_{2} = \frac{k Q}{r_{2}}$ since, $r_{1} \neq r_{2}$
$V_{1} \neq V_{2}$ are consequently there will be a potential difference between the two conductors.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359415 Assertion :
For practical purposes, the earth is used as a reference at zero potential in electrical circuits.
Reason :
The electrical potential of a sphere of radius $R$ with charge $Q$ uniformly distributed on the surface is given by $\frac{Q}{4 π ε_{0} R}$ . As $R$ is very large and $Q$ is very small (total charge of earth) so the net potential of earth is approximately zero.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359416 The potential of a large liquid drop when eight liquid drops are combined is $20 V$ . Then the potential of each single drop was

1

10 V

2

7.5 V

3

5 V

4

2.5 V

Explanation:

Let $R$ be radius of the big drop and $r$ be radius of each small drop.
$As volume of the big drop = volume of$
$eight small drops$
$∴ \frac{4}{3} π R^{3} = 8 (\frac{4}{3} π r^{3})$
$R = 2 r (1)$
If $q$ be charge on each small drop, then charge on the big drop is
$Q = 8 q (2)$
The potential of each small drop is $V = \frac{1}{4 π ε_{0}} \frac{q}{r} (3)$
and that of the big drop is
$V^{'} = \frac{1}{4 π ε_{0}} \frac{Q}{R} (4)$
Dividing eqn. (3) by eqn. (4), we get
$\frac{V}{V^{'}} = \frac{\frac{1}{4 π ε_{0}} \frac{q}{r}}{\frac{1}{4 π ε_{0}} \frac{Q}{R}} = \frac{q}{Q} \frac{R}{r}$
$= \frac{q}{8 q} \frac{2 r}{r} = \frac{1}{4} (Using eqns. (1) and (2))$
$V = \frac{1}{4} V^{'}$
As $V^{'} = 20 V$ (given)
$∴ V = \frac{1}{4} (20 V) = 5 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359417 A hollow conducting sphere is placed in an electric field produced by a point charge placed at $P$ as shown in figure. Let $V_{A}, V_{B}, V_{C}$ be the potential at points $A$ , $B$ and $C$ respectively. Then
supporting img

1

V_{B} > V_{C}

2

V_{A} > V_{B}

3

V_{C} > V_{B}

4

V_{A} = V_{C}