(a) The part of the area of the plate with the dielectric K at any time is bx. Its capacitance is C1=Kε0bxd
Similarly, the capacitance of the part without the dielectric is C2=ε0b(ℓ−x)d
These two parts are connected in parallel. The capitance of the system is therefore,
C=C1+C2=ε0bd[ℓ+x(K−1)] ...(1)
The energy of the capacitor is
U=12CV2=ε0bV22d[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 is therefore,
dWb=VdQ=(dC)V2
The external force F does a work
dWe=(−Fdx)
during the displacement. The total work done on the capacitor is
dWb+dWe=(dC)V2−Fdx.
This should be equal to the increase dU in the stored energy. Thus,
12(dC)V2=(dC)V2−Fdx
F=12V2(dC/dx)=ε0bV2[K−1]2d
Thus, the electric field attracts the dielectric into the capacitor with a force ε0bV2[K−1]2d and this minimum force should be applied in the opposite direction, so as to remove it from the capacitor.