Question

Figure (31-E30) shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width b and lengths l₁ and l₂. The left half of the dielectric slab has a dielectric constant K₁ and the right half K₂. Neglecting any friction, find the ration of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium.

Answer 1

Let the potential of the battery connected to the left capacitor be V₁ and that of the battery connected to the right capacitor be V_2.

Considering the left capacitor,
Let the length of the part of the slab inside the capacitor be x.

The left capacitor can be considered to be two capacitors in parallel.

The capacitances of the two capacitors in parallel are:
$C_1=\frac{K_1{\in }_{0}bx}{d},C_2=\frac{{\in }_{0}b(l_1-x)}{d}$
C₁ is the part of the capacitor having the dielectric inserted in it and C₂ is the capacitance of the part of the capacitor without dielectric.

As, C₁ and C₂ are in parallel.

Therefore, the net capacitance is given by
C = C₁ + C₂
$$\begin{gathered} C=\frac{K_1{\in }_{0}bx}{d}+\frac{{\in }_{0}b(l_1-x)}{d} \\ C=\frac{{\in }_{0}b}{d}\left[l_1+x(K_1-1)\right] \end{gathered}$$


Therefore, the potential energy stored in the left capacitor will be
$$\begin{gathered} U=\frac{1}{2}C{V}_1^2 \\ U=\frac{{\in }_{0}b{V}_1^2}{2d}\left[l_1+x(K_1-1)\right]...\left(1\right) \end{gathered}$$

The dielectric slab is attracted by the electric field of the capacitor and applies a force in left direction.
Let us consider electric force of magnitude F pulls the slab in left direction.
Let there be an infinitesimal displacement dx in left direction by the force F.
The work done by the force = F.dx


Let us consider a small displacement dx of the slab in the inward direction. The capacitance will increase, therefore the energy of the capacitor will also increase. In order to maintain constant voltage, the battery will supply extra charges, therefore the battery will do work.

Work done by the battery = change in energy of capacitor + work done by the force F on the capacitor
dW_B = dU + dW_F
Let the charge dq is supplied by the battery, and the change in the capacitor be dC
dW_B = (dq).V = (dC).V²
dU = $\frac{1}{2}$(dC).V²

(dC).V² = $\frac{1}{2}$(dC).V² + F.dx
$\frac{1}{2}$(dC).V² = F.dx
$$\begin{gathered} \Rightarrow F=\frac{1}{2}\frac{dC}{dx}{V}_1^2 \\ \Rightarrow F=\frac{1}{2}\frac{d}{dx}\left(\frac{{\in }_{0}b}{d}\left[l_1+x(K_1-1)\right]\right)V_1^2 \\ \Rightarrow F=\frac{{\in }_{0}b{V}_1^2(K_1-1)}{2d} \\ \Rightarrow V_1^2=\frac{F\times 2d}{{\in }_{0}b(K_1-1)}\Rightarrow V_1=\sqrt{\frac{F\times 2d}{{\in }_{0}b(K_1-1)}} \end{gathered}$$

Similarly, for the right side the voltage of the battery is given by
$V_2=\sqrt{\frac{F\times 2d}{{\in }_{0}b(K_2-1)}}$

Thus, the ratio of the voltages is given by
$$\begin{gathered} \frac{V_1}{V_2}=\frac{\sqrt{\frac{F\times 2d}{{\in }_{0}b({\mathrm{K}}_1-1)}}}{\sqrt{\frac{F\times 2d}{{\in }_{0}b({\mathrm{K}}_2-1)}}} \\ \Rightarrow \frac{V_1}{V_2}=\frac{\sqrt{K_2-1}}{\sqrt{K_1-1}} \end{gathered}$$

Thus, the ratio of the emfs of the left battery to the right battery is $\frac{V_1}{V_2}=\frac{\sqrt{K_2-1}}{\sqrt{K_1-1}}$