(B) Initially, the charge on the $10\,\mu F$ capacitor is:$Q = C_1 V_1 = (10\,\mu F)(50\, V) = 500\,\mu C$When this capacitor is connected in parallel

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(B) Initially, the charge on the $10\,\mu F$ capacitor is:$Q = C_1 V_1 = (10\,\mu F)(50\, V) = 500\,\mu C$When this capacitor is connected in parallel to an uncharged capacitor of capacitance $C_2$, the total charge remains conserved.Let the final potential difference be $V = 20\, V$.The total charge $Q$ is distributed between the two capacitors:$Q = (C_1 + C_2)V$$500\,\mu C = (10\,\mu F + C_2)(20\, V)$Dividing both sides by $20\, V$:$25\,\mu F = 10\,\mu F + C_2$$C_2 = 25\,\mu F - 10\,\mu F = 15\,\mu F$

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

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$A$ $10\,\mu F$ capacitor is fully charged to a potential difference of $50\, V$. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now, the potential difference across them becomes $20\, V$. The capacitance of the second capacitor is $\dots \mu F$.

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$15$
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$20$
D
$30$

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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$A$ $10\,\mu F$ capacitor is fully charged to a potential difference of $50\, V$. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now, the potential difference across them becomes $20\, V$. The capacitance of the second capacitor is $\dots \mu F$.

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$A$ $20 \text{ F}$ capacitor is charged to $5 \text{ V}$ and isolated. It is then connected in parallel with an uncharged $30 \text{ F}$ capacitor. The decrease in the energy of the system will be: (in $\text{ J}$)

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