JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Capacitance

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2 μ 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}

8 μ F

. The charge in

C_{2}

on equilibrium condition is ______

μ C

. (Round off to the Nearest integer)

Answer: 16

Solution

Initial charge on

C_{1}

Q_{0} = C_{1} V = 2 \times 10 = 20

μC. When connected to uncharged

C_{2}

, charge redistributes. At equilibrium, both have the same potential

V^{'}

. By charge conservation:

Q_{0} = Q_{1}^{'} + Q_{2}^{'} = C_{1} V^{'} + C_{2} V^{'} = (C_{1} + C_{2}) V^{'}

. Therefore

V^{'} = \frac{Q _{0}}{C _{1} + C _{2}} = \frac{20}{2 + 8} = 2

V. Charge on

C_{2}

Q_{2}^{'} = C_{2} V^{'} = 8 \times 2 = 16

μC.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer

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