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Consider the situation shown in figure. The plates of the capacitor have plate area A and are clamped in the laboratory. the dielectric slab is released from rest with a length a inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period.

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$8 \sqrt{\frac{(l + a) md}{ε_{0} {Aε}^{2} (K + 1)}}$

$8 \sqrt{\frac{(l - a) lmd}{ε_{0} {Aε}^{2} (K - 1)}}$

$8 \sqrt{\frac{(l - a) lmd}{ε_{0} {Aε}^{2} (K + 1)}}$

$8 \sqrt{\frac{(l + a) lmd}{ε_{0} {Aε}^{2} (K + 1)}}$

answer is A.

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Detailed Solution

$\begin{array}{l} F = \frac{ε_{0} {bV}^{2} (K - 1)}{2 d}; \\ a = \frac{F}{m} \Rightarrow a = \frac{ε_{0} {AV}^{2} (K - 1)}{2 mℓd} \end{array}$

At the instant when the slab is fully inside the plates, we get the state of equilibrium for the slab. So, time taken to reach from the given position to equilibrium position equals t (say)

$s = \frac{1}{2} {at}^{2} \Rightarrow t = \sqrt{\frac{2 s}{a}} (∵ s = ℓ - x);$

$t = \sqrt{\frac{2 s}{a}} = \sqrt{\frac{2 (l - x)}{\frac{ε_{0} {AV}^{2} (K - 1)}{2 mℓd}}}$

$T = 4 t = 8 \sqrt{\frac{(l - x) mℓd}{ε_{0} {AV}^{2} (K - 1)}}$

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