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Capacitance

165589 The distance between two plates of a capacitor is $d$ and its capacitance is $C_{1}$ , when air is the medium between the plates. If a metal sheet of thickness $\frac{2 d}{3}$ and of same area as plate is introduced between the plates, the capacitance of the capacitor becomes $C_{2}$ . The ratio $\frac{C_{2}}{C_{1}}$ is:

1

2 : 1

2

4 : 1

3

3 : 1

4

1 : 1

Explanation:

: Given,
Case 1 - When air is the medium between the plates.
$Capacitance (C_{1}) = \frac{ε_{o} A}{d}$
Case 2 - When, $t = \frac{2 d}{3}$ thickness sheet introduce between plates
$Capacitance (C_{2}) = \frac{ε_{o}}{d - t + \frac{t}{K}}$
$C_{2} = \frac{ε_{o} A}{d - \frac{2 d}{3} + \frac{t}{\infty}}$
$C_{2} = \frac{ε_{o} A}{\frac{d}{3}}$
$C_{2} = 3 \frac{ε_{0} A}{d}$
Ratio of capacitance $C_{1}$ and $C_{2}$
$\frac{C_{2}}{C_{1}} = \frac{3 \frac{ε_{0} A}{d}}{\frac{ε_{0} A}{d}} = \frac{3}{1}$

[JEE Main-10.04.2023

Capacitance

165591 A big drop of radius $R$ is formed by 729 small drops of water of radius $r$ , then the radius of each small drop will be:

1

\frac{R}{9}

2

\frac{R}{900}

3

\frac{R}{1800}

4

\frac{R}{9000}

Explanation:

: Number of drops (n) $= 729$
Radius of big drop $= R$
Radius of each small drop $= r$
Equating volume in big drop and small drop,
$\frac{4}{3} π R^{3} = n \frac{4}{3} π r^{3}$
$R^{3} = 729 r^{3}$
$r = \frac{R^{3}}{(729)^{1 / 3}}$
$r = \frac{R}{9}$

[AIIMS-1997]

Capacitance

165592 The earth is assumed to be a charged conducting sphere having volume ' $V$ ' and surface area ' $A$ '. The capacitance of the earth in free space is
( $ε_{0} =$ permittivity of free space)

1

4 π ε_{0} \frac{V}{A}

2

8 π ε_{0} \frac{V}{A}

3

2 π ε_{0} \frac{V}{A}

4

12 π ε_{0} \frac{V}{A}

Explanation:

: Volume of earth $(V) = \frac{4}{3} π R^{3}$
Area of earth $(A) = 4 π R^{2}$
Ratio of volume of earth to the area of earth,
$\frac{V}{A} = \frac{\frac{4}{3} π R^{3}}{4 π R^{2}}$
$\frac{V}{A} = \frac{R}{3} \Rightarrow R = \frac{3 V}{A}$
Where, $R \to$ radius of earth
$∵$ Capacitance of earth in free space,
$C = 4 π ε_{0} R$
$= 4 π ε_{0} \times \frac{3 V}{A}$
$C = 12 π ε_{o} \times \frac{V}{A}$

[MHT-CET 2020]

Capacitance

165594 Assume each oil drop consists of a capacitance of $C$ . If combine $n$ drops to form a bigger drop, then the capacitance of bigger drop $C^{'}$ would be

1

C^{'} = \frac{2 n^{1 / 3}}{3} C

2

C^{'} = \frac{5 n^{1 / 3}}{4} C

3

C^{'} = \frac{n^{1 / 3}}{5} C

4

C^{'} = C \cdot n^{1 / 3}

Explanation:

: Radius of oil drop $= r$
Capacitance of an oil drop $(C) = 4 π ε_{0} r$
$∵ n \times$ volume of each small drop $=$ volume of large drop
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$R = n^{1 / 3} . r$
So, capacitance of large drop,
$C^{'} = 4 π ε_{0} R = 4 π ε_{0} n^{1 / 3} r$
$C^{'} = n^{1 / 3} C$

[TS- EAMCET-10.09.2020

Capacitance

165596 A capacitor of capacitance $10 μ F$ is charged to potential $50 V$ with a battery. The battery is now disconnected and an additional charge $200 μ C$ is given to the positive plate of the capacitor. The potential difference across the capacitor will be

1

50 V

2

80 V

3

100 V

4

60 V

Explanation:

: Given, Capacitor of capacitance (C) $= 10 μ F$ Charged to potential $(V) = 50 V$

Charge $(Q) = CV$
$= 10 \times 50 = 500 μ C$
$∵$ Additional charge to the positive plate $(Q^{'}) = 200 μ C$

Total charge on positive plate $= 700 μ C$
Total charge in negative plate $= - 500 μ C$
$∵$ Net electric field at point $P$ is zero
$E_{p} = 0$

$∴ \frac{700 - Q}{2 A ε_{o}} + \frac{Q}{2 A ε_{o}} + \frac{500 - Q}{2 A ε_{o}} = \frac{Q}{2 A ε_{o}}$
$Q = 600 μ C$
$∴$ Potential difference between the plates,
$Δ V = \frac{Q}{C} = \frac{600}{10} = 60 V$

[Manipal UGET-2015]

Capacitance

165589 The distance between two plates of a capacitor is $d$ and its capacitance is $C_{1}$ , when air is the medium between the plates. If a metal sheet of thickness $\frac{2 d}{3}$ and of same area as plate is introduced between the plates, the capacitance of the capacitor becomes $C_{2}$ . The ratio $\frac{C_{2}}{C_{1}}$ is:

1

2 : 1

2

4 : 1

3

3 : 1

4

1 : 1

Explanation:

: Given,
Case 1 - When air is the medium between the plates.
$Capacitance (C_{1}) = \frac{ε_{o} A}{d}$
Case 2 - When, $t = \frac{2 d}{3}$ thickness sheet introduce between plates
$Capacitance (C_{2}) = \frac{ε_{o}}{d - t + \frac{t}{K}}$
$C_{2} = \frac{ε_{o} A}{d - \frac{2 d}{3} + \frac{t}{\infty}}$
$C_{2} = \frac{ε_{o} A}{\frac{d}{3}}$
$C_{2} = 3 \frac{ε_{0} A}{d}$
Ratio of capacitance $C_{1}$ and $C_{2}$
$\frac{C_{2}}{C_{1}} = \frac{3 \frac{ε_{0} A}{d}}{\frac{ε_{0} A}{d}} = \frac{3}{1}$

[JEE Main-10.04.2023

Capacitance

165591 A big drop of radius $R$ is formed by 729 small drops of water of radius $r$ , then the radius of each small drop will be:

1

\frac{R}{9}

2

\frac{R}{900}

3

\frac{R}{1800}

4

\frac{R}{9000}

Explanation:

: Number of drops (n) $= 729$
Radius of big drop $= R$
Radius of each small drop $= r$
Equating volume in big drop and small drop,
$\frac{4}{3} π R^{3} = n \frac{4}{3} π r^{3}$
$R^{3} = 729 r^{3}$
$r = \frac{R^{3}}{(729)^{1 / 3}}$
$r = \frac{R}{9}$

[AIIMS-1997]

Capacitance

165592 The earth is assumed to be a charged conducting sphere having volume ' $V$ ' and surface area ' $A$ '. The capacitance of the earth in free space is
( $ε_{0} =$ permittivity of free space)

1

4 π ε_{0} \frac{V}{A}

2

8 π ε_{0} \frac{V}{A}

3

2 π ε_{0} \frac{V}{A}

4

12 π ε_{0} \frac{V}{A}

Explanation:

: Volume of earth $(V) = \frac{4}{3} π R^{3}$
Area of earth $(A) = 4 π R^{2}$
Ratio of volume of earth to the area of earth,
$\frac{V}{A} = \frac{\frac{4}{3} π R^{3}}{4 π R^{2}}$
$\frac{V}{A} = \frac{R}{3} \Rightarrow R = \frac{3 V}{A}$
Where, $R \to$ radius of earth
$∵$ Capacitance of earth in free space,
$C = 4 π ε_{0} R$
$= 4 π ε_{0} \times \frac{3 V}{A}$
$C = 12 π ε_{o} \times \frac{V}{A}$

[MHT-CET 2020]

Capacitance

165594 Assume each oil drop consists of a capacitance of $C$ . If combine $n$ drops to form a bigger drop, then the capacitance of bigger drop $C^{'}$ would be

1

C^{'} = \frac{2 n^{1 / 3}}{3} C

2

C^{'} = \frac{5 n^{1 / 3}}{4} C

3

C^{'} = \frac{n^{1 / 3}}{5} C

4

C^{'} = C \cdot n^{1 / 3}

Explanation:

: Radius of oil drop $= r$
Capacitance of an oil drop $(C) = 4 π ε_{0} r$
$∵ n \times$ volume of each small drop $=$ volume of large drop
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$R = n^{1 / 3} . r$
So, capacitance of large drop,
$C^{'} = 4 π ε_{0} R = 4 π ε_{0} n^{1 / 3} r$
$C^{'} = n^{1 / 3} C$

[TS- EAMCET-10.09.2020

Capacitance

165596 A capacitor of capacitance $10 μ F$ is charged to potential $50 V$ with a battery. The battery is now disconnected and an additional charge $200 μ C$ is given to the positive plate of the capacitor. The potential difference across the capacitor will be

1

50 V

2

80 V

3

100 V

4

60 V

Explanation:

: Given, Capacitor of capacitance (C) $= 10 μ F$ Charged to potential $(V) = 50 V$

Charge $(Q) = CV$
$= 10 \times 50 = 500 μ C$
$∵$ Additional charge to the positive plate $(Q^{'}) = 200 μ C$

Total charge on positive plate $= 700 μ C$
Total charge in negative plate $= - 500 μ C$
$∵$ Net electric field at point $P$ is zero
$E_{p} = 0$

$∴ \frac{700 - Q}{2 A ε_{o}} + \frac{Q}{2 A ε_{o}} + \frac{500 - Q}{2 A ε_{o}} = \frac{Q}{2 A ε_{o}}$
$Q = 600 μ C$
$∴$ Potential difference between the plates,
$Δ V = \frac{Q}{C} = \frac{600}{10} = 60 V$

[Manipal UGET-2015]

Capacitance

165589 The distance between two plates of a capacitor is $d$ and its capacitance is $C_{1}$ , when air is the medium between the plates. If a metal sheet of thickness $\frac{2 d}{3}$ and of same area as plate is introduced between the plates, the capacitance of the capacitor becomes $C_{2}$ . The ratio $\frac{C_{2}}{C_{1}}$ is:

1

2 : 1

2

4 : 1

3

3 : 1

4

1 : 1

Explanation:

: Given,
Case 1 - When air is the medium between the plates.
$Capacitance (C_{1}) = \frac{ε_{o} A}{d}$
Case 2 - When, $t = \frac{2 d}{3}$ thickness sheet introduce between plates
$Capacitance (C_{2}) = \frac{ε_{o}}{d - t + \frac{t}{K}}$
$C_{2} = \frac{ε_{o} A}{d - \frac{2 d}{3} + \frac{t}{\infty}}$
$C_{2} = \frac{ε_{o} A}{\frac{d}{3}}$
$C_{2} = 3 \frac{ε_{0} A}{d}$
Ratio of capacitance $C_{1}$ and $C_{2}$
$\frac{C_{2}}{C_{1}} = \frac{3 \frac{ε_{0} A}{d}}{\frac{ε_{0} A}{d}} = \frac{3}{1}$

[JEE Main-10.04.2023

Capacitance

165591 A big drop of radius $R$ is formed by 729 small drops of water of radius $r$ , then the radius of each small drop will be:

1

\frac{R}{9}

2

\frac{R}{900}

3

\frac{R}{1800}

4

\frac{R}{9000}

Explanation:

: Number of drops (n) $= 729$
Radius of big drop $= R$
Radius of each small drop $= r$
Equating volume in big drop and small drop,
$\frac{4}{3} π R^{3} = n \frac{4}{3} π r^{3}$
$R^{3} = 729 r^{3}$
$r = \frac{R^{3}}{(729)^{1 / 3}}$
$r = \frac{R}{9}$

[AIIMS-1997]

Capacitance

165592 The earth is assumed to be a charged conducting sphere having volume ' $V$ ' and surface area ' $A$ '. The capacitance of the earth in free space is
( $ε_{0} =$ permittivity of free space)

1

4 π ε_{0} \frac{V}{A}

2

8 π ε_{0} \frac{V}{A}

3

2 π ε_{0} \frac{V}{A}

4

12 π ε_{0} \frac{V}{A}

Explanation:

: Volume of earth $(V) = \frac{4}{3} π R^{3}$
Area of earth $(A) = 4 π R^{2}$
Ratio of volume of earth to the area of earth,
$\frac{V}{A} = \frac{\frac{4}{3} π R^{3}}{4 π R^{2}}$
$\frac{V}{A} = \frac{R}{3} \Rightarrow R = \frac{3 V}{A}$
Where, $R \to$ radius of earth
$∵$ Capacitance of earth in free space,
$C = 4 π ε_{0} R$
$= 4 π ε_{0} \times \frac{3 V}{A}$
$C = 12 π ε_{o} \times \frac{V}{A}$

[MHT-CET 2020]

Capacitance

165594 Assume each oil drop consists of a capacitance of $C$ . If combine $n$ drops to form a bigger drop, then the capacitance of bigger drop $C^{'}$ would be

1

C^{'} = \frac{2 n^{1 / 3}}{3} C

2

C^{'} = \frac{5 n^{1 / 3}}{4} C

3

C^{'} = \frac{n^{1 / 3}}{5} C

4

C^{'} = C \cdot n^{1 / 3}

Explanation:

: Radius of oil drop $= r$
Capacitance of an oil drop $(C) = 4 π ε_{0} r$
$∵ n \times$ volume of each small drop $=$ volume of large drop
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$R = n^{1 / 3} . r$
So, capacitance of large drop,
$C^{'} = 4 π ε_{0} R = 4 π ε_{0} n^{1 / 3} r$
$C^{'} = n^{1 / 3} C$

[TS- EAMCET-10.09.2020

Capacitance

165596 A capacitor of capacitance $10 μ F$ is charged to potential $50 V$ with a battery. The battery is now disconnected and an additional charge $200 μ C$ is given to the positive plate of the capacitor. The potential difference across the capacitor will be

1

50 V

2

80 V

3

100 V

4

60 V

Explanation:

: Given, Capacitor of capacitance (C) $= 10 μ F$ Charged to potential $(V) = 50 V$

Charge $(Q) = CV$
$= 10 \times 50 = 500 μ C$
$∵$ Additional charge to the positive plate $(Q^{'}) = 200 μ C$

Total charge on positive plate $= 700 μ C$
Total charge in negative plate $= - 500 μ C$
$∵$ Net electric field at point $P$ is zero
$E_{p} = 0$

$∴ \frac{700 - Q}{2 A ε_{o}} + \frac{Q}{2 A ε_{o}} + \frac{500 - Q}{2 A ε_{o}} = \frac{Q}{2 A ε_{o}}$
$Q = 600 μ C$
$∴$ Potential difference between the plates,
$Δ V = \frac{Q}{C} = \frac{600}{10} = 60 V$

[Manipal UGET-2015]

Capacitance

165589 The distance between two plates of a capacitor is $d$ and its capacitance is $C_{1}$ , when air is the medium between the plates. If a metal sheet of thickness $\frac{2 d}{3}$ and of same area as plate is introduced between the plates, the capacitance of the capacitor becomes $C_{2}$ . The ratio $\frac{C_{2}}{C_{1}}$ is:

1

2 : 1

2

4 : 1

3

3 : 1

4

1 : 1

Explanation:

: Given,
Case 1 - When air is the medium between the plates.
$Capacitance (C_{1}) = \frac{ε_{o} A}{d}$
Case 2 - When, $t = \frac{2 d}{3}$ thickness sheet introduce between plates
$Capacitance (C_{2}) = \frac{ε_{o}}{d - t + \frac{t}{K}}$
$C_{2} = \frac{ε_{o} A}{d - \frac{2 d}{3} + \frac{t}{\infty}}$
$C_{2} = \frac{ε_{o} A}{\frac{d}{3}}$
$C_{2} = 3 \frac{ε_{0} A}{d}$
Ratio of capacitance $C_{1}$ and $C_{2}$
$\frac{C_{2}}{C_{1}} = \frac{3 \frac{ε_{0} A}{d}}{\frac{ε_{0} A}{d}} = \frac{3}{1}$

[JEE Main-10.04.2023

Capacitance

165591 A big drop of radius $R$ is formed by 729 small drops of water of radius $r$ , then the radius of each small drop will be:

1

\frac{R}{9}

2

\frac{R}{900}

3

\frac{R}{1800}

4

\frac{R}{9000}

Explanation:

: Number of drops (n) $= 729$
Radius of big drop $= R$
Radius of each small drop $= r$
Equating volume in big drop and small drop,
$\frac{4}{3} π R^{3} = n \frac{4}{3} π r^{3}$
$R^{3} = 729 r^{3}$
$r = \frac{R^{3}}{(729)^{1 / 3}}$
$r = \frac{R}{9}$

[AIIMS-1997]

Capacitance

165592 The earth is assumed to be a charged conducting sphere having volume ' $V$ ' and surface area ' $A$ '. The capacitance of the earth in free space is
( $ε_{0} =$ permittivity of free space)

1

4 π ε_{0} \frac{V}{A}

2

8 π ε_{0} \frac{V}{A}

3

2 π ε_{0} \frac{V}{A}

4

12 π ε_{0} \frac{V}{A}

Explanation:

: Volume of earth $(V) = \frac{4}{3} π R^{3}$
Area of earth $(A) = 4 π R^{2}$
Ratio of volume of earth to the area of earth,
$\frac{V}{A} = \frac{\frac{4}{3} π R^{3}}{4 π R^{2}}$
$\frac{V}{A} = \frac{R}{3} \Rightarrow R = \frac{3 V}{A}$
Where, $R \to$ radius of earth
$∵$ Capacitance of earth in free space,
$C = 4 π ε_{0} R$
$= 4 π ε_{0} \times \frac{3 V}{A}$
$C = 12 π ε_{o} \times \frac{V}{A}$

[MHT-CET 2020]

Capacitance

165594 Assume each oil drop consists of a capacitance of $C$ . If combine $n$ drops to form a bigger drop, then the capacitance of bigger drop $C^{'}$ would be

1

C^{'} = \frac{2 n^{1 / 3}}{3} C

2

C^{'} = \frac{5 n^{1 / 3}}{4} C

3

C^{'} = \frac{n^{1 / 3}}{5} C

4

C^{'} = C \cdot n^{1 / 3}

Explanation:

: Radius of oil drop $= r$
Capacitance of an oil drop $(C) = 4 π ε_{0} r$
$∵ n \times$ volume of each small drop $=$ volume of large drop
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$R = n^{1 / 3} . r$
So, capacitance of large drop,
$C^{'} = 4 π ε_{0} R = 4 π ε_{0} n^{1 / 3} r$
$C^{'} = n^{1 / 3} C$

[TS- EAMCET-10.09.2020

Capacitance

165596 A capacitor of capacitance $10 μ F$ is charged to potential $50 V$ with a battery. The battery is now disconnected and an additional charge $200 μ C$ is given to the positive plate of the capacitor. The potential difference across the capacitor will be

1

50 V

2

80 V

3

100 V

4

60 V

Explanation:

: Given, Capacitor of capacitance (C) $= 10 μ F$ Charged to potential $(V) = 50 V$

Charge $(Q) = CV$
$= 10 \times 50 = 500 μ C$
$∵$ Additional charge to the positive plate $(Q^{'}) = 200 μ C$

Total charge on positive plate $= 700 μ C$
Total charge in negative plate $= - 500 μ C$
$∵$ Net electric field at point $P$ is zero
$E_{p} = 0$

$∴ \frac{700 - Q}{2 A ε_{o}} + \frac{Q}{2 A ε_{o}} + \frac{500 - Q}{2 A ε_{o}} = \frac{Q}{2 A ε_{o}}$
$Q = 600 μ C$
$∴$ Potential difference between the plates,
$Δ V = \frac{Q}{C} = \frac{600}{10} = 60 V$

[Manipal UGET-2015]

Capacitance

165589 The distance between two plates of a capacitor is $d$ and its capacitance is $C_{1}$ , when air is the medium between the plates. If a metal sheet of thickness $\frac{2 d}{3}$ and of same area as plate is introduced between the plates, the capacitance of the capacitor becomes $C_{2}$ . The ratio $\frac{C_{2}}{C_{1}}$ is:

1

2 : 1

2

4 : 1

3

3 : 1

4

1 : 1

Explanation:

: Given,
Case 1 - When air is the medium between the plates.
$Capacitance (C_{1}) = \frac{ε_{o} A}{d}$
Case 2 - When, $t = \frac{2 d}{3}$ thickness sheet introduce between plates
$Capacitance (C_{2}) = \frac{ε_{o}}{d - t + \frac{t}{K}}$
$C_{2} = \frac{ε_{o} A}{d - \frac{2 d}{3} + \frac{t}{\infty}}$
$C_{2} = \frac{ε_{o} A}{\frac{d}{3}}$
$C_{2} = 3 \frac{ε_{0} A}{d}$
Ratio of capacitance $C_{1}$ and $C_{2}$
$\frac{C_{2}}{C_{1}} = \frac{3 \frac{ε_{0} A}{d}}{\frac{ε_{0} A}{d}} = \frac{3}{1}$

[JEE Main-10.04.2023

Capacitance

165591 A big drop of radius $R$ is formed by 729 small drops of water of radius $r$ , then the radius of each small drop will be:

1

\frac{R}{9}

2

\frac{R}{900}

3

\frac{R}{1800}

4

\frac{R}{9000}

Explanation:

: Number of drops (n) $= 729$
Radius of big drop $= R$
Radius of each small drop $= r$
Equating volume in big drop and small drop,
$\frac{4}{3} π R^{3} = n \frac{4}{3} π r^{3}$
$R^{3} = 729 r^{3}$
$r = \frac{R^{3}}{(729)^{1 / 3}}$
$r = \frac{R}{9}$

[AIIMS-1997]

Capacitance

165592 The earth is assumed to be a charged conducting sphere having volume ' $V$ ' and surface area ' $A$ '. The capacitance of the earth in free space is
( $ε_{0} =$ permittivity of free space)

1

4 π ε_{0} \frac{V}{A}

2

8 π ε_{0} \frac{V}{A}

3

2 π ε_{0} \frac{V}{A}

4

12 π ε_{0} \frac{V}{A}

Explanation:

: Volume of earth $(V) = \frac{4}{3} π R^{3}$
Area of earth $(A) = 4 π R^{2}$
Ratio of volume of earth to the area of earth,
$\frac{V}{A} = \frac{\frac{4}{3} π R^{3}}{4 π R^{2}}$
$\frac{V}{A} = \frac{R}{3} \Rightarrow R = \frac{3 V}{A}$
Where, $R \to$ radius of earth
$∵$ Capacitance of earth in free space,
$C = 4 π ε_{0} R$
$= 4 π ε_{0} \times \frac{3 V}{A}$
$C = 12 π ε_{o} \times \frac{V}{A}$

[MHT-CET 2020]

Capacitance

165594 Assume each oil drop consists of a capacitance of $C$ . If combine $n$ drops to form a bigger drop, then the capacitance of bigger drop $C^{'}$ would be

1

C^{'} = \frac{2 n^{1 / 3}}{3} C

2

C^{'} = \frac{5 n^{1 / 3}}{4} C

3

C^{'} = \frac{n^{1 / 3}}{5} C

4

C^{'} = C \cdot n^{1 / 3}

Explanation:

: Radius of oil drop $= r$
Capacitance of an oil drop $(C) = 4 π ε_{0} r$
$∵ n \times$ volume of each small drop $=$ volume of large drop
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$R = n^{1 / 3} . r$
So, capacitance of large drop,
$C^{'} = 4 π ε_{0} R = 4 π ε_{0} n^{1 / 3} r$
$C^{'} = n^{1 / 3} C$

[TS- EAMCET-10.09.2020

Capacitance

165596 A capacitor of capacitance $10 μ F$ is charged to potential $50 V$ with a battery. The battery is now disconnected and an additional charge $200 μ C$ is given to the positive plate of the capacitor. The potential difference across the capacitor will be

1

50 V

2

80 V

3

100 V

4

60 V

Explanation:

: Given, Capacitor of capacitance (C) $= 10 μ F$ Charged to potential $(V) = 50 V$

Charge $(Q) = CV$
$= 10 \times 50 = 500 μ C$
$∵$ Additional charge to the positive plate $(Q^{'}) = 200 μ C$

Total charge on positive plate $= 700 μ C$
Total charge in negative plate $= - 500 μ C$
$∵$ Net electric field at point $P$ is zero
$E_{p} = 0$

$∴ \frac{700 - Q}{2 A ε_{o}} + \frac{Q}{2 A ε_{o}} + \frac{500 - Q}{2 A ε_{o}} = \frac{Q}{2 A ε_{o}}$
$Q = 600 μ C$
$∴$ Potential difference between the plates,
$Δ V = \frac{Q}{C} = \frac{600}{10} = 60 V$

[Manipal UGET-2015]