Figure shows two parallel plate capacitors with fixed plates and … - Vidyadip

Question

Figure 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 $l_1$ and $l_2$ The left half of the dielectric slab has a dielectric constant $K_1$ and the right half $K_2$. Neglecting any 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.

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Answer

Consider the left side
The plate area of the part with the dielectric is by its capacitance $\text{C}_1=\frac{\text{K}_1\epsilon_0\text{bx}}{\text{d}}$ and with out dielectric $\text{C}_2=\frac{\epsilon_0\text{b}(\text{L}_1-\text{x})}{\text{d}}$
These are connected in parallel
$\text{C}=\text{C}_1+\text{C}_2$
$=\frac{\epsilon_0\text{b}}{\text{d}}[\text{L}_1+\text{x}(\text{k}_1-1)]$
Let the potential $V_1$
$\text{U}=\Big(\frac{1}{2}\Big)\text{CV}_1^2=\frac{\epsilon_0\text{bv}_1^2}{2\text{d}}[\text{L}_1+\text{x}(\text{k}-1)]\ \dots(1)$
Suppose dielectric slab is attracted by electric field and an external force $F$ consider the part dx which makes inside further, As the potential difference remains constant at $V.$
The charge supply, $dq = (dc)v$ to the capacitor.
The work done by the battery is $dw_b = v.dq = (dc)v^2$
The external force $F$ does a work dwe $= (–f.dx)$ during a small displacement.
The total work done in the capacitor is $dw_b + dw_e = (dc)v^2 - fdx$
This should be equal to the increase dv in the stored energy.
Thus $\Big(\frac{1}{2}\Big)(\text{dk})\text{v}^2=(\text{dc})\text{v}^2-\text{fdx}$
$\text{f}=\frac{1}{2}\text{v}^2\frac{\text{dc}}{\text{dx}}$
from equation $(1)$
$\text{F}=\frac{\epsilon_0\text{bv}^2}{2\text{d}}(\text{k}_1-1)$
$\Rightarrow\text{V}_1^2=\frac{\text{F}\times2\text{d}}{\epsilon_0\text{b}(\text{k}_1-1)}$
$\Rightarrow\text{V}_1=\sqrt{\frac{\text{F}\times2\text{d}}{\epsilon_0\text{b}(\text{k}_1-1)}}$
For the right side, $\text{V}_2=\sqrt{\frac{\text{F}\times2\text{d}}{\epsilon_0\text{b}(\text{k}_2-1)}}$
$\frac{\text{V}_1}{\text{V}_2}=\frac{\sqrt{\frac{\text{F}\times2\text{d}}{\epsilon_0\text{b}(\text{k}_1-1)}}}{\sqrt{\frac{\text{F}\times2\text{d}}{\epsilon_0\text{b}(\text{k}_2-1)}}}$
$\frac{\text{V}_1}{\text{V}_2}=\frac{\sqrt{\text{k}_2-1}}{\sqrt{\text{k}_1-1}}$
$\therefore$ The ratio of the emf of the left battery to the right battery $=\frac{\sqrt{\text{k}_2-1}}{\sqrt{\text{k}_1-1}}$

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