JEE PYQ: Capacitance Question 18
Question 18 - 2020 (05 Sep Shift 2)
A parallel plate capacitor has plate of length ‘$l$’, width ‘$w$’ and separation of plates is ‘$d$’. It is connected to a battery of emf V. A dielectric slab of the same thickness ‘$d$’ and of dielectric constant $k = 4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?
(1) $2l/3$
(2) $l/3$
(3) $l/4$
(4) $l/2$
Show Answer
Answer: (2)
Solution
Capacitance before inserting slab: $C_i = \frac{\epsilon_0 wl}{d}$. After inserting length $x$: $C_f = \frac{K\epsilon_0 wx}{d} + \frac{\epsilon_0 w(l-x)}{d}$. $2 \times \frac{1}{2}C_i V^2 = \frac{1}{2}C_f V^2$. $2C_i = C_f$. $2\frac{\epsilon_0 wl}{d} = \frac{\epsilon_0 w}{d}[kx + (l-x)]$. $2l = 4x + l - x = 3x + l$. $x = \frac{l}{3}$.
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