Two square plates of sides $l$ are placed parallel to each other with separation $d$ as suggested in figure. You may assume $d$ is much less than $l$ . The plates carry uniformly distributed static charges $+ Q_{0}$ and $- Q_{0}$ . A block of metal has width $l$ , length $l$ , and thickness slightly less than $d$ and it is inserted at distance $x$ into the space between the plates. The charges on the plates remain uniformly distributed as the block slides in. In a static situation, a metal prevents an electrical field from penetrating inside it. The metal can be thought of as a perfect dielectric, with $K \to \infty$

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The area of the advancing front face of the block is essentially equal to $l d$ , the stress(force per unit area) on it is $\frac{{Q_{0}}^{2}}{(2 l^{4} \in_{0})}$

The stored energy as a function of x is given by $\frac{{Q_{0}}^{2} d (l - x)}{(2 l^{3} \in_{0})}$

The energy density in the electric field between the charged plates in terms of $ Q_{0,} l, d$ and $\in_{0}$ is $\frac{{Q_{0}}^{2}}{(2 l^{4} \in_{0})}$

The magnitude of the force that acts on the metallic block is given by $\frac{{Q_{0}}^{2} d}{(l^{3} \in_{0})}$ to the right

answer is A, C, D.

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Detailed Solution

$U = \frac{1}{2} \in_{0} E^{2} \times v o l u m e = \frac{1}{2} \in_{0} {(\frac{Q_{0}}{l^{2} \in_{0}})}^{2} (l - x) l d = \frac{1}{2} \frac{{Q_{0}}^{2}}{l^{3} \in_{0}} (l - x) d F = \frac{- d U}{d X} = - [\frac{1}{2} \frac{{Q_{0}}^{2}}{l^{3} \in_{0}} (- 1) d] = \frac{{Q_{0}}^{2} d}{2 l^{3} \in_{0}} H e n c e \frac{F}{A} = \frac{F}{l d} = \frac{{Q_{0}}^{2}}{2 l^{4} \in_{0}}$