A parallel plate capacitor has plate of length 'l', width ‘w’ and… | JEE Main 2020 (5 Sep, Shift 2)

Q90JEE Main 2020 · 5 SEP · SHIFT 2Difficult

Dielectrics

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?

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Answer: (d)

Battery connected, so $V$ is constant and $U = \frac{1}{2}CV^2 \propto C$ ; need $C' = 2C_0$ .
With slab length $x$ inside (parallel parts): $C' = \frac{\epsilon_0 w}{d}[Kx + (l - x)] = \frac{\epsilon_0 w}{d}(l + 3x)$ .
$l + 3x = 2l \Rightarrow x = \frac{l}{3}$
So the answer is (d) $\frac{l}{3}$ .

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