Figure shows a parallel-plate capacitor with plates of width b and length l. The separation between the plates is d. The plates are rigidly clamped and connected to a battery of emf V. A dielectric slab of thickness d and dielectric constant K is slowly inserted between the plates. Calculate the energy of the system when a length x of the slab is introduced into the capacitor.

Explanation

(a) The plates area of the part with the dielectric at any time is bx. Its capacitance is `C_(1)=(Kepsilon_(0)bx)/(d)`
Similarly, the capacitance of the part without the dielectric is
`C_(2)=(epsilon_(0)b(l-x))/(d)`
These two parts are connected in parallel. The capacitance of the system is, therefore,
`C=C_(1)+C_(2)=(epsilon_(0)b)/(d)[l+x(K-1)]" "...(i)`
The energy of the capacitor is
`U=(1)/(2)CV^(2)=(epsilon_(0)bV^(2))/(2d)[l+x(K-1)]`
(b) Suppose, the electric field attracts the dielectric slab with a force F. An external force of equal magnitude F should be applied in opposite direction so that the plate moves slowly (no acceleration).
Consider the part of motion in which the dielectric moves a distance dx further inside the capacitor. The capacitance increases to `C+dC`. As the potential difference remains constant at V, the battery has to supply a further charge
`dQ=(dC)V`
to the capacitor. The work done by the battery is, therefore,
`dW_(b)=VdQ=(dC)V^(2)`
The external force F does a work
`dW_(e)=(-Fdx)`
during the displacemen. The total work done on the capacitor is
`dW_(b)+dW_(e)=(dC)V^(2)-Fdx`.
This should be equal to the increase dU in the stored energy. Thus,
`(1)/(2)(dC)V^(2)=(dC)V^(2)-Fdx`
`F=(1)/(2)V^(2)(dC//dx)=(epsilon_(0)bV^(2)[K-1])/(2d)`
Thus, the electric field attracts the dielectric into the capacitor with a force `(epsilon_(0)bV^(2)[K-1])/(2d)` and this minimum force should be applied in the opposite direction, so as to remove it from the capacitor.