Question

Figure shows a parallel plate capacitor with plates of width $b$ and length $\ell$. 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.
(a) Calculate the energy of the system when a length $x$ of the slab is introduced into the capacitor.
(b) What force should be applied on the slab to ensure that goes slowly into the capacitor? Neglect any effect of friction or gravity.

Answer 1

(a) The part of the area of the plate with the dielectric $K$ at any time is $bx$. Its capacitance is $C_1=\frac{K{\epsilon }_{0}bx}{d}$
Similarly, the capacitance of the part without the dielectric is $C_2=\frac{{\epsilon }_{0}b(\ell -x)}{d}$
These two parts are connected in parallel. The capitance of the system is therefore,
$C=C_1+C_2=\frac{{\epsilon }_{0}b}{d}[\ell +x(K-1\left)\right]$ ...(1)
The energy of the capacitor is
$U=\frac{1}{2}C{V}^2=\frac{{\epsilon }_{0}b{V}^2}{2d}[\ell +x(K-1\left)\right]$
(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,
$d{W}_b=VdQ=\left(dC\right)V^2$
The external force $F$ does a work
$d{W}_e=(-Fdx)$
during the displacement. The total work done on the capacitor is
$d{W}_b+d{W}_e=\left(dC\right)V^2-Fdx$.
This should be equal to the increase $dU$ in the stored energy. Thus,
$\frac{1}{2}\left(dC\right)V^2=\left(dC\right)V^2-Fdx$
$F=\frac{1}{2}{V}^2\left(dC/dx\right)=\frac{{\epsilon }_{0}b{V}^2[K-1]}{2d}$
Thus, the electric field attracts the dielectric into the capacitor with a force $\frac{{\epsilon }_{0}b{V}^2[K-1]}{2d}$ and this minimum force should be applied in the opposite direction, so as to remove it from the capacitor.