A parallel plate capacitor has plates of length $l$, width $w$ and separation $d$. 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 the plates will the energy stored in the capacitor be two times the initial energy stored?
Answer: (B) $\dfrac{l}{3}$
With the battery connected, $U = \tfrac12 CV^2$, so the energy doubles when the capacitance doubles.
With a length $x$ of the slab inside, the two parts act as capacitors in parallel:
$$C = \frac{\varepsilon_0 w}{d}\left[(l-x) + Kx\right] = C_0\left(1 + \frac{3x}{l}\right)$$
$$1 + \frac{3x}{l} = 2 \Rightarrow x = \frac{l}{3}$$
Solution by Sreeraj P, M.Sc Physics