Question

A parallel plate capacitor of plate area A and plate separation d is charged to the potential difference V and then the battery is 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 electric field between the plates (after the slab is inserted) and work done on the system in question in the process of inserting the slab, then
(A) $\mathrm{Q}=\frac{{\mathrm{\epsilon }}_0\mathrm{AV}}{\mathrm{d}}$ (B) $\mathrm{Q}=\frac{{\mathrm{\epsilon }}_0\mathrm{KAV}}{\mathrm{d}}$ (C) $\mathrm{E}=\frac{\mathrm{V}}{\mathrm{Kd}}$ (D) $\mathrm{W}=\frac{{\mathrm{\epsilon }}_0{\mathrm{AV}}^2}{2\mathrm{d}}\left[1-\frac{1}{\mathrm{K}}\right]$

Answer 1

Dear Student,


Work done (W) on the system during the process of inserting a dielectric slab is stored as the potential energy (U) of the
system and is given as


$U=\frac{1}{2}C{V}^2\left(1\right),whereC=\frac{{\epsilon }_{0}KA}{d}$

Substituting for C in equation (1), we get

$U=\frac{{\epsilon }_{0}KA{V}^2}{2d}\left(2\right)$
U can also be expressed in terms of charge density Q and potential difference V as


$U=\frac{1}{2}QV\left(3\right)$

Comparing equations (2) and (3), we get the expression for charge density as

$Q=\frac{{\epsilon }_{0}KAV}{d}$

Thus option (B) is the right choice





Regards