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Capacitance

165590 Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason R.
Assertion A: Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one.
Reason R: Capacitance of metallic spheres depend on the radii of spheres.
In the light of the above statements, choose the correct answer from the options given below.

1 Both

A

and

R

are true and

R

is correct explanation of

A

2 Both

A

and

R

are true but

R

is not the correct explanation of

A

3 A is true but

R

is false

4 A is false but

R

is true

[JEE Main-01.02.2023

Capacitance

165593 A capacitor of unknown capacitance $C$ is connected across a battery of $V$ volt. The charge stored in it becomes $Q$ coulomb. When potential across the capacitor is reduced by $V^{'}$ volt, the charge stored in it becomes $Q^{'}$ coulomb. The capacitance $C$ is

1

\frac{Q - Q^{'}}{V^{'}}

2

\frac{V^{'}}{Q - Q^{'}}

3

\frac{Q + Q^{'}}{V^{'}}

4

\frac{Q - Q^{'}}{\sqrt{V^{'}}}

Explanation:

: We know that -
$Q = CV$
According to the question,
$Q^{'} = C (V - V^{'}) = CV - {CV}^{'}$
$Q^{'} = Q - {CV}^{'}$
$C V^{'} = Q - Q^{'}$
$C = \frac{Q - Q^{'}}{V^{'}}$

[MHT-CET 2020]

Capacitance

165595 When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then

1 charge on capacitor increases

2 potential difference across the capacitor increases

3 energy stored increases

4 capacity remains the same

Explanation:

: Capacitor of a conductor is ratio of charge (Q) to it by rise in its potential (V).
$C = \frac{Q}{V} = \frac{k ε_{0} A}{d}$
$∴$ When a dielectric slab is introduced between the plates of a capacitor connected to battery, Thus capacity of capacitor will increase i.e., charge on capacitor increases and voltage across capacitor decreases.

[Manipal UGET -2020]

Capacitance

165626 A capacitor of capacitance $C_{1}$ is charged to a potential $V$ and then connected in parallel to an uncharged capacitor of capacitance $C_{2}$ . The final potential difference across each capacitor will be

1

\frac{C_{1} V}{C_{1} + C_{2}}

2

\frac{C_{2} V}{C_{1} + C_{2}}

3

1 + \frac{C_{2}}{C_{1}}

4

1 - \frac{C_{2}}{C_{1}}

Explanation:

: The common potential difference across two capacitors connected in parallel,
$V_{eq} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$
Here, potential of charged capacitor,
Potential of uncharged capacitor $(V_{2}) = 0$
$∴ V_{eq} = \frac{C_{1} V}{C_{1} + C_{2}}$

[J and K CET- 2011]

Capacitance

165590 Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason R.
Assertion A: Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one.
Reason R: Capacitance of metallic spheres depend on the radii of spheres.
In the light of the above statements, choose the correct answer from the options given below.

1 Both

A

and

R

are true and

R

is correct explanation of

A

2 Both

A

and

R

are true but

R

is not the correct explanation of

A

3 A is true but

R

is false

4 A is false but

R

is true

[JEE Main-01.02.2023

Capacitance

165593 A capacitor of unknown capacitance $C$ is connected across a battery of $V$ volt. The charge stored in it becomes $Q$ coulomb. When potential across the capacitor is reduced by $V^{'}$ volt, the charge stored in it becomes $Q^{'}$ coulomb. The capacitance $C$ is

1

\frac{Q - Q^{'}}{V^{'}}

2

\frac{V^{'}}{Q - Q^{'}}

3

\frac{Q + Q^{'}}{V^{'}}

4

\frac{Q - Q^{'}}{\sqrt{V^{'}}}

Explanation:

: We know that -
$Q = CV$
According to the question,
$Q^{'} = C (V - V^{'}) = CV - {CV}^{'}$
$Q^{'} = Q - {CV}^{'}$
$C V^{'} = Q - Q^{'}$
$C = \frac{Q - Q^{'}}{V^{'}}$

[MHT-CET 2020]

Capacitance

165595 When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then

1 charge on capacitor increases

2 potential difference across the capacitor increases

3 energy stored increases

4 capacity remains the same

Explanation:

: Capacitor of a conductor is ratio of charge (Q) to it by rise in its potential (V).
$C = \frac{Q}{V} = \frac{k ε_{0} A}{d}$
$∴$ When a dielectric slab is introduced between the plates of a capacitor connected to battery, Thus capacity of capacitor will increase i.e., charge on capacitor increases and voltage across capacitor decreases.

[Manipal UGET -2020]

Capacitance

165626 A capacitor of capacitance $C_{1}$ is charged to a potential $V$ and then connected in parallel to an uncharged capacitor of capacitance $C_{2}$ . The final potential difference across each capacitor will be

1

\frac{C_{1} V}{C_{1} + C_{2}}

2

\frac{C_{2} V}{C_{1} + C_{2}}

3

1 + \frac{C_{2}}{C_{1}}

4

1 - \frac{C_{2}}{C_{1}}

Explanation:

: The common potential difference across two capacitors connected in parallel,
$V_{eq} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$
Here, potential of charged capacitor,
Potential of uncharged capacitor $(V_{2}) = 0$
$∴ V_{eq} = \frac{C_{1} V}{C_{1} + C_{2}}$

[J and K CET- 2011]

Capacitance

165590 Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason R.
Assertion A: Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one.
Reason R: Capacitance of metallic spheres depend on the radii of spheres.
In the light of the above statements, choose the correct answer from the options given below.

1 Both

A

and

R

are true and

R

is correct explanation of

A

2 Both

A

and

R

are true but

R

is not the correct explanation of

A

3 A is true but

R

is false

4 A is false but

R

is true

[JEE Main-01.02.2023

Capacitance

165593 A capacitor of unknown capacitance $C$ is connected across a battery of $V$ volt. The charge stored in it becomes $Q$ coulomb. When potential across the capacitor is reduced by $V^{'}$ volt, the charge stored in it becomes $Q^{'}$ coulomb. The capacitance $C$ is

1

\frac{Q - Q^{'}}{V^{'}}

2

\frac{V^{'}}{Q - Q^{'}}

3

\frac{Q + Q^{'}}{V^{'}}

4

\frac{Q - Q^{'}}{\sqrt{V^{'}}}

Explanation:

: We know that -
$Q = CV$
According to the question,
$Q^{'} = C (V - V^{'}) = CV - {CV}^{'}$
$Q^{'} = Q - {CV}^{'}$
$C V^{'} = Q - Q^{'}$
$C = \frac{Q - Q^{'}}{V^{'}}$

[MHT-CET 2020]

Capacitance

165595 When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then

1 charge on capacitor increases

2 potential difference across the capacitor increases

3 energy stored increases

4 capacity remains the same

Explanation:

: Capacitor of a conductor is ratio of charge (Q) to it by rise in its potential (V).
$C = \frac{Q}{V} = \frac{k ε_{0} A}{d}$
$∴$ When a dielectric slab is introduced between the plates of a capacitor connected to battery, Thus capacity of capacitor will increase i.e., charge on capacitor increases and voltage across capacitor decreases.

[Manipal UGET -2020]

Capacitance

165626 A capacitor of capacitance $C_{1}$ is charged to a potential $V$ and then connected in parallel to an uncharged capacitor of capacitance $C_{2}$ . The final potential difference across each capacitor will be

1

\frac{C_{1} V}{C_{1} + C_{2}}

2

\frac{C_{2} V}{C_{1} + C_{2}}

3

1 + \frac{C_{2}}{C_{1}}

4

1 - \frac{C_{2}}{C_{1}}

Explanation:

: The common potential difference across two capacitors connected in parallel,
$V_{eq} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$
Here, potential of charged capacitor,
Potential of uncharged capacitor $(V_{2}) = 0$
$∴ V_{eq} = \frac{C_{1} V}{C_{1} + C_{2}}$

[J and K CET- 2011]

Capacitance

165590 Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason R.
Assertion A: Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one.
Reason R: Capacitance of metallic spheres depend on the radii of spheres.
In the light of the above statements, choose the correct answer from the options given below.

1 Both

A

and

R

are true and

R

is correct explanation of

A

2 Both

A

and

R

are true but

R

is not the correct explanation of

A

3 A is true but

R

is false

4 A is false but

R

is true

[JEE Main-01.02.2023

Capacitance

165593 A capacitor of unknown capacitance $C$ is connected across a battery of $V$ volt. The charge stored in it becomes $Q$ coulomb. When potential across the capacitor is reduced by $V^{'}$ volt, the charge stored in it becomes $Q^{'}$ coulomb. The capacitance $C$ is

1

\frac{Q - Q^{'}}{V^{'}}

2

\frac{V^{'}}{Q - Q^{'}}

3

\frac{Q + Q^{'}}{V^{'}}

4

\frac{Q - Q^{'}}{\sqrt{V^{'}}}

Explanation:

: We know that -
$Q = CV$
According to the question,
$Q^{'} = C (V - V^{'}) = CV - {CV}^{'}$
$Q^{'} = Q - {CV}^{'}$
$C V^{'} = Q - Q^{'}$
$C = \frac{Q - Q^{'}}{V^{'}}$

[MHT-CET 2020]

Capacitance

165595 When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then

1 charge on capacitor increases

2 potential difference across the capacitor increases

3 energy stored increases

4 capacity remains the same

Explanation:

: Capacitor of a conductor is ratio of charge (Q) to it by rise in its potential (V).
$C = \frac{Q}{V} = \frac{k ε_{0} A}{d}$
$∴$ When a dielectric slab is introduced between the plates of a capacitor connected to battery, Thus capacity of capacitor will increase i.e., charge on capacitor increases and voltage across capacitor decreases.

[Manipal UGET -2020]

Capacitance

165626 A capacitor of capacitance $C_{1}$ is charged to a potential $V$ and then connected in parallel to an uncharged capacitor of capacitance $C_{2}$ . The final potential difference across each capacitor will be

1

\frac{C_{1} V}{C_{1} + C_{2}}

2

\frac{C_{2} V}{C_{1} + C_{2}}

3

1 + \frac{C_{2}}{C_{1}}

4

1 - \frac{C_{2}}{C_{1}}

Explanation:

: The common potential difference across two capacitors connected in parallel,
$V_{eq} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$
Here, potential of charged capacitor,
Potential of uncharged capacitor $(V_{2}) = 0$
$∴ V_{eq} = \frac{C_{1} V}{C_{1} + C_{2}}$

[J and K CET- 2011]

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