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Consider the situation shown in figure. The width of each plate is b and distance between the plates is d. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf E. All surfaces are frictionless. Calculate the value of M for which the dielectric slab will stay in equilibrium.

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$\frac{ε_{0} {bE}^{2} (K - 1)}{2 dg}$

$\frac{ε_{0} {bE}^{2} (K - 1)}{dg}$

$\frac{2 ε_{0} {bE}^{2} (K - 1)}{dg}$

$\frac{ε_{0} {bE}^{2} (K + 1)}{dg}$

answer is A.

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Detailed Solution

$∵$ electric force on the slab

$F = \frac{du}{dx} = \frac{1}{2} \frac{ε_{0} {bE}^{2}}{d} (K - 1)$

$Mg = \frac{ε_{0} {bE}^{2}}{2 d} (K - 1)$

$M = \frac{ε_{0} {bE}^{2}}{2 dg} (K - 1)$

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