Consider the left side The plate area of the part with the dielectric is by its capacitance $\text{C}_1=\frac{\text{K}_1\in_0\text{bx}}{\text{d}}$ and with out dielectric $\text{C}_2=\frac{\in_0\text{b}(\text{L}_1-\text{x})}{\text{d}}$ These are connected in parallel $\text{C}=\text{C}_1+\text{C}_2$
$=\frac{\in_0\text{b}}{\text{d}}[\text{L}_1+\text{x}(\text{k}_1-1)]$
Let the potential V1$\text{U}=\Big(\frac{1}{2}\Big)\text{CV}_1^2=\frac{\in_0\text{bv}_1^2}{2\text{d}}[\text{L}_1+\text{x}(\text{k}-1)]\ \dots(1)$
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 dwb = v.dq = (dc)v2 The external force F does a work dwe = (–f.dx) during a small displacement. The total work done in the capacitor is dwb + dwe = (dc)v2 - 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}$$\text{f}=\frac{1}{2}\text{v}^2\frac{\text{dc}}{\text{dx}}$
from equation (1)$\text{F}=\frac{\in_0\text{bv}^2}{2\text{d}}(\text{k}_1-1)$
$\Rightarrow\text{V}_1^2=\frac{\text{F}\times2\text{d}}{\in_0\text{b}(\text{k}_1-1)}$
$\Rightarrow\text{V}_1=\sqrt{\frac{\text{F}\times2\text{d}}{\in_0\text{b}(\text{k}_1-1)}}$
For the right side, $\text{V}_2=\sqrt{\frac{\text{F}\times2\text{d}}{\in_0\text{b}(\text{k}_2-1)}}$$\frac{\text{V}_1}{\text{V}_2}=\frac{\sqrt{\frac{\text{F}\times2\text{d}}{\in_0\text{b}(\text{k}_1-1)}}}{\sqrt{\frac{\text{F}\times2\text{d}}{\in_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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