A 2 μ F capacitor C 1 is first charged to a potential difference of 10   V using a battery. Then…

A 2μF capacitor C1 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 C2 of 8μF. The charge in C2 on equilibrium condition is μC. (Round off to the Nearest Integer)

Solution

20=C1+C2 VV=2 volt .

Q2=C2V=16 μC

=16

Asked in: JEE Main 2021 (17 Mar Shift 2)

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