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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance5 Marks
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 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 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 $\text{C}_1=\frac{\text{K}_1\in_0\text{bx}}{\text{d}}$ 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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