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JEE Main202117 Mar 2021Evening ShiftPhysicsCapacitanceActual

A 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 of 8 μ F . The charge in C 2 on equilibrium condition is μ C . (Round off to the Nearest Integer)

Correct answer

0

Step-by-step solution

20 = C 1 + C 2   V ⇒ V = 2   volt   . ⇒ Q 2 = C 2 V = 16   μ C = 16

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