(D) $1$. Initial charge on capacitor $C_{1}$ is $Q = C_{1}V = 2\, \mu F \times 10\, V = 20\, \mu C$.$2$. When the battery is removed and $C_{1}$ is co

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(D) $1$. Initial charge on capacitor $C_{1}$ is $Q = C_{1}V = 2\, \mu F \times 10\, V = 20\, \mu C$.$2$. When the battery is removed and $C_{1}$ is connected to the uncharged capacitor $C_{2}$,the total charge $Q$ is conserved and shared between the two capacitors.$3$. The common potential $V'$ after connection is given by $V' = \frac{Q}{C_{1} + C_{2}} = \frac{20\, \mu C}{2\, \mu F + 8\, \mu F} = \frac{20}{10} = 2\, V$.$4$. The charge on capacitor $C_{2}$ at equilibrium is $Q_{2} = C_{2}V' = 8\, \mu F \times 2\, V = 16\, \mu C$.

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

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$A$ $2\, \mu F$ capacitor $C_{1}$ is first charged to a potential difference of $10\, V$ using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor $C_{2}$ of $8\, \mu F$. The charge in $C_{2}$ on equilibrium condition is $\ldots\, \mu C$. (Round off to the Nearest Integer)

JEE MAIN 2021 All JEE MAIN

A
$9$
B
$25$
C
$20$
D
$16$

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Class 12

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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Two metallic spheres of radii $1\,cm$ and $2\,cm$ are given charges $10^{-2}\,C$ and $5 \times 10^{-2}\,C$ respectively. If they are connected by a conducting wire,the final charge on the smaller sphere is

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