A parallel plate capacitor of plate area A & plate separation d is charged to a potential difference V & then the battery disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of the charge on each plate, the magnitude of the electric field between the plates (after the slab is inserted) & the magnitude of the work done on the system, in the process of inserting the slab, then :

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E = $\frac{V}{K d}$

W = $\frac{\in_{0} A V^{2}}{2 d} (1 - \frac{1}{K})$

Q = $\frac{\in_{0} A V}{d}$

Q = $\frac{\in_{0} K A V}{d}$

answer is A, C, D.

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Detailed Solution

(A, C, D)
C = $\frac{\in_{0} A}{d}$ , C' = $\frac{K \in_{0} A}{d}$ Q = CV = $\frac{\in_{0} K A V}{d}$
$Q = C V = C_{1} V_{1}$ $\Rightarrow$ $V_{1} = \frac{V}{K}$ E = $\frac{V_{1}}{d} = \frac{V}{K d}$
$W = U_{f} - U_{i} = C V^{2} - \frac{1}{2} C_{1} {V_{1}}^{2} = \frac{1}{2} \frac{\in_{0} A V^{2}}{d^{2}} - \frac{1}{2} \frac{K \in_{0} A}{d} {(\frac{V}{K})}^{2} = \frac{\in_{0} ⥄ A V^{2}}{2 d} (1 - \frac{1}{K})$