JEE PYQ: Capacitance Question 1
Question 1 - 2021 (16 Mar Shift 1)
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4}d$, where ’d’ is the separation between the plates of parallel plate capacitor. The new capacitance (C’) in terms of original capacitance ($C_0$) is given by the following relation:
(1) $C’ = \frac{3+K}{4K}C_0$
(2) $C’ = \frac{4+K}{3}C_0$
(3) $C’ = \frac{4K}{K+3}C_0$
(4) $C’ = \frac{4}{3+K}C_0$
Show Answer
Answer: (3)
Solution
$C_0 = \frac{\epsilon_0 A}{d}$. $C’ = C_1$ and $C_2$ in series. $\frac{1}{C’} = \frac{3d/4}{\epsilon_0 KA} + \frac{d/4}{\epsilon_0 A}$. $\frac{1}{C’} = \frac{d}{4\epsilon_0 A}\left(\frac{3+K}{K}\right)$. $C’ = \frac{4KC_0}{(3+K)}$.
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