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

The ratio of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium is $=\frac{\sqrt{\mathrm{k}_{2}-1}}{\sqrt{\mathrm{k}_{1}-1}}$

Explanation:

Consider the left side,

Capacitance $C_{1}=\frac{k_{1} \varepsilon_{0} b x}{d}$

and with out dielectric, $C_{2}=\frac{\varepsilon_{0} b\left(L_{1}-x\right)}{d}$

These are connected in parallel,

$\mathrm{c}=\mathrm{c}_{1}+\mathrm{C}_{2}=\frac{\varepsilon_{0} \mathrm{b}}{\mathrm{d}}\left[\mathrm{L}_{1}+\mathrm{x}\left(\mathrm{k}_{1}-1\right)\right]$

Let the potential $V_1$

$U=(1 / 2) C V_{1}^{2}=\frac{\varepsilon_{0} b v_{1}^{2}}{2 d}\left[L_{1}+x(k-1)\right]$           ................(1)

$\text { Thus }(1 / 2)(d k) v^{2}=(d c) v^{2}-f d x$

                         $f=\frac{1}{2} v^{2} \frac{d c}{d x}$

from equation (1)

$\mathrm{F}=\frac{\varepsilon_{0} \mathrm{bv}^{2}}{2 \mathrm{d}}\left(\mathrm{k}_{1}-1\right)$

$V_{1}^{2}=\frac{F \times 2 d}{\varepsilon_{0} b\left(k_{1}-1\right)}$

$V_{1}=\sqrt{\frac{F \times 2 d}{\varepsilon_{0} b\left(k_{1}-1\right)}}$

For the right side, $V_{2}=\sqrt{\frac{F \times 2 d}{\varepsilon_{0} b\left(k_{2}-1\right)}}$

$\frac{v_{1}}{v_{2}}=\frac{\sqrt{\frac{F \times 2 d}{\varepsilon_{0} b\left(k_{1}-1\right)}}}{\sqrt{\frac{F \times 2 d}{\varepsilon_{0} b\left(k_{2}-1\right)}}}$

$\frac{V_{1}}{V_{2}}=\frac{\sqrt{k_{2}-1}}{\sqrt{k_{1}-1}}$

Hence the ratio is   $=\frac{\sqrt{k_{2}-1}}{\sqrt{k_{1}-1}}$

Here we can represent the variables as

C = Capacitance of the circuit

A = Area of the plates of the capacitor

d = distance between two capacitor plates

k= dielectric constant of the material that used between the two plates of the capacitor. This dielectric constant is acts as a medium between the plates which helps to pass the conductivity between the plates.