A 2 μ 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 F. The charge in C2 on equilibrium condition is ____________ μ C. (Round off to the Nearest Integer) Numerical answer type. Work it out, then open the answer.

Answer: 16. Q = 2 × 10 = 20 μ C Common potential V = (20)/(2 + 8) = 2 V q₂ = 8 × 2 = 16 μ C

A 2 μ F capacitor C1 is first charged to a potential difference of… | JEE Main 2021 (17 Mar, Shift 2)

Q23JEE Main 2021 · 17 MAR · SHIFT 2Easy

Charge Sharing & Loss of Energy

A 2 $\mu$ 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 $\mu$ F. The charge in C2 on equilibrium condition is ____________ $\mu$ C. (Round off to the Nearest Integer)

Numerical answer type. Work it out, then open the answer.

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Answer: 16

$Q = 2 \times 10 = 20\ \mu$ C
Common potential $V = \frac{20}{2 + 8} = 2$ V
$q_2 = 8 \times 2 = 16\ \mu$ C

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Charge Sharing & Loss of Energy question from JEE Main 2021 (17 Mar, Shift 2)

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