(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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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

Two parallel plate capacitors of capacitances $20 \mu F$ and $30 \mu F$ are charged to potentials of $30 \, V$ and $20 \, V$ respectively. If the similarly charged plates are connected together,then the common potential difference will be: (in $, V$)

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$A$ $2\,\mu F$ capacitor is charged to a potential of $10\,V$. Another $4\,\mu F$ capacitor is charged to a potential of $20\,V$. The two capacitors are then connected in a single loop,with the positive plate of one connected to the negative plate of the other. What heat is evolved in the circuit in $\mu J$?

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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

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Two parallel plate capacitors of capacitances $20 \mu F$ and $30 \mu F$ are charged to potentials of $30 \, V$ and $20 \, V$ respectively. If the similarly charged plates are connected together,then the common potential difference will be: (in $, V$)

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