Question

Figure shows two parallel plate capacitors with fixed plates and connected to two batteries.The separation between the plates is the same for two capacitors. The plates are rectangular in shape with width $b$ and lengths $l_1$ and $l_2$. The left half of the dielectric slab has a dielectric constant $K_1$ and the right half $K_2$. Neglecting friction, find the ratio of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium. [Take $K_1=2$ and $K_2=5$]

Answer 1

(a) Consider the left side:
The plate area of the left part with the dielectric is $bn$.
Its capacitance, $C_1=\frac{K_1{\epsilon }_{0}bn}{d}$

And without dielectric $C_2=\frac{{\epsilon }_{0}b(L_1-n)}{d}$

These are connected in parallel, so the equivalent capacitance,

$C=C_1+C_2=\frac{{\epsilon }_{0}b}{d}[L_1+n(K_1-1\left)\right].......\left(1\right)$

Let the potential difference or emf of the battery $V_1$.

Suppose, dielectric slab is attracted by electric field and an external force $F$.

Consider the part $dn$ which makes inside further, As the potential differences remains constant as $V$.

The charge supply, $dq=\left(dC\right)V$ to the capacitor.

The work done by the battery is $d{W}_b=Vdq=\left(dC\right)V^2$

The external force $F$ does a work $d{W}_e=(-F.dn)$ during a small displacement.

The total work done will be
$d{W}_b+d{W}_e=\left(dC\right)V^2-Fdn$

This should be equal to the increase $dV$ in the stored energy.

Thus, $\frac{1}{2}\left(dC\right)V^2=\left(dC\right)V^2-Fdn$

$\Rightarrow F=\frac{1}{2}{V}^2\frac{dC}{dn}$

From equation (1)
$F=\frac{{\epsilon }_{0}b{V}^2}{2d}(K_1-1)$

$\Rightarrow V_1^2=\frac{F\times 2d}{{\epsilon }_{0}b(K_1-1)}$

$\Rightarrow V_1=\sqrt{\frac{F\times 2d}{{\epsilon }_{0}b(K_1-1)}}$

Similarly, for right side potential difference,

$\Rightarrow V_2=\sqrt{\frac{F\times 2d}{{\epsilon }_{0}b(K_2-1)}}$

$\therefore \frac{V_1}{V_2}=\frac{\sqrt{\frac{F\times 2d}{{\epsilon }_{0}b(K_1-1)}}}{\sqrt{\frac{F\times 2d}{{\epsilon }_{0}b(K_2-1)}}}$

$\Rightarrow \frac{V_1}{V_2}=\frac{\sqrt{K_2-1}}{\sqrt{K_1-1}}$

Given, $K_1=2$ and $K_2=5$

$\Rightarrow \frac{V_1}{V_2}=\frac{\sqrt{5-1}}{\sqrt{2-1}}=2$

Accepted answer: 2, 2.0, 2.00