(C) The capacitor $C$ is initially charged to a potential difference $V_0$. The initial charge on the capacitor is $Q = CV_0$.When this capacitor is c

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$C\left( \frac{V_0}{V} - 1 \right)$

(C) The capacitor $C$ is initially charged to a potential difference $V_0$. The initial charge on the capacitor is $Q = CV_0$.When this capacitor is connected to an uncharged capacitor of capacitance $C_x$,the total charge $Q$ is conserved and distributed between the two capacitors.Since they are connected in parallel,they will share a common potential difference $V$.The total charge remains $Q = (C + C_x)V$.Equating the initial and final charges: $CV_0 = (C + C_x)V$.Rearranging for $C_x$: $C + C_x = \frac{CV_0}{V}$.Therefore,$C_x = C\left( \frac{V_0}{V} - 1 \right)$.

Class 12 Physics Electric Potential and Capacitance Sharing of Charge in Capacitor Circuit

Medium

$A$ capacitor of capacitance $C$ is charged to a potential difference $V_0$. The charging battery is removed and the capacitor is now connected to an uncharged capacitor of unknown capacity. The potential difference across the combination becomes $V$. The unknown capacitance is

A
$C\left( \frac{V_0}{V} \right)$
B
$C\left( \frac{V_0}{V} + 1 \right)$
C
$C\left( \frac{V_0}{V} - 1 \right)$
D
$CV$

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Class 12

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel plate capacitor charged to a potential $V$. The two capacitors share charges and the common potential becomes $V^{\prime}$. The dielectric constant $K$ is

Medium

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Two circuits $(a)$ and $(b)$ have charged capacitors with capacitances and charges as shown in the figures. The switches are initially open. On closing the switches, what happens to the charge flow?

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Circuits $(a)$ and $(b)$ have charged capacitors with capacitances $C, 2C$ and $3C$ as shown,with an open switch $S$. When the switch is closed,determine the direction of charge flow for each circuit.

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Using a battery,a $100 \ pF$ capacitor is charged to $60 \ V$ and then the battery is removed. After that,a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20 \ V$,its capacitance is: (in $pF$)

MediumJEE MAIN 2025

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Two conducting spheres of radii $R_1$ and $R_2$ are charged with charges $Q_1$ and $Q_2$ respectively. When they are brought into contact with each other:

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$A$ capacitor of capacitance $C$ is charged to a potential difference $V_0$. The charging battery is removed and the capacitor is now connected to an uncharged capacitor of unknown capacity. The potential difference across the combination becomes $V$. The unknown capacitance is

Similar Questions

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel plate capacitor charged to a potential $V$. The two capacitors share charges and the common potential becomes $V^{\prime}$. The dielectric constant $K$ is

Two circuits $(a)$ and $(b)$ have charged capacitors with capacitances and charges as shown in the figures. The switches are initially open. On closing the switches, what happens to the charge flow?

Circuits $(a)$ and $(b)$ have charged capacitors with capacitances $C, 2C$ and $3C$ as shown,with an open switch $S$. When the switch is closed,determine the direction of charge flow for each circuit.

Using a battery,a $100 \ pF$ capacitor is charged to $60 \ V$ and then the battery is removed. After that,a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20 \ V$,its capacitance is: (in $pF$)

Two conducting spheres of radii $R_1$ and $R_2$ are charged with charges $Q_1$ and $Q_2$ respectively. When they are brought into contact with each other:

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