Electrostatic Potential and Capacitance - Result Question 38
40. A capacitor $C_1$ is charged to a potential difference $V$. The charging battery is then removed and the capacitor is connected to an uncharged capacitor $C_2$. The potential difference across the combination is
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(a) $\frac{V C_1}{(C_1+C_2)}$
(b) $V(1+\frac{C_2}{C_1})$
(c) $V(1+\frac{C_1}{C_2})$
(d) $\frac{V C_2}{(C_1+C_2)}$
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Answer:
Correct Answer: 40. (a)
Solution:
- (a) Charge $Q=C_1 V$
Total capacity of combination (parallel)
$C=C_1+C_2 \quad$ P.D. $=\frac{Q}{C}=\frac{C_1 V}{C_1+C_2}$
