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QJEEM21Ph3a6e: JEE Main 2021 Physics - Class 12 JEEM21Ph3a6e Solution: A 2 F capacitor C 1 is first charged to a potential difference…

A 2 μ\mu F capacitor C1 is first charged to a potential difference of 10V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C2 of 8 μ\mu F. The charge in C2 on equilibrium condition is ____________ μ\mu C. (Round off to the Nearest Integer)

JEE Main 2021 (Online) 17th March Evening Shift Physics - Capacitor Question 85 English

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JEE Main 2021 (Online) 17th March Evening Shift Physics - Capacitor Question 85 English Explanation
When battery is removed & the capacitor is connected

2V + 8v = 20

10V = 20

V = 2 volt

∵\because Q = CV

Q = 8 ×\times 2 = 16 μ\mu c

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