JEE PYQ: Capacitance Question 28
Question 28 - 2019 (09 Apr Shift 1)
A capacitor with capacitance 5 $\mu$F is charged to 5 $\mu$C. If the plates are pulled apart to reduce the capacitance to 2 $\mu$F, how much work is done?
(1) $6.25 \times 10^{-6}$ J
(2) $3.75 \times 10^{-6}$ J
(3) $2.16 \times 10^{-6}$ J
(4) $2.55 \times 10^{-6}$ J
Show Answer
Answer: (2)
Solution
$W = U_f - U_i = \frac{q^2}{2}\left(\frac{1}{C_f} - \frac{1}{C_i}\right) = \frac{(5 \times 10^{-6})^2}{2}\left(\frac{1}{2} - \frac{1}{5}\right) \times 10^6 = 3.75 \times 10^{-6}$ J.
BETA
