A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant and , as shwn in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are

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Solution

Q.
A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $ε_{1}$ and $ε_{2}$ , as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are $C_{1}$ and $C_{2}$ respectively, then $\frac{C_{1}}{C_{2}}$ is [2025]

1 $\frac{ε_{1} ε_{2}^{2}}{{(ε_{1} + ε_{2})}^{2}}$

2 $\frac{4 ε_{1} ε_{2}}{{(ε_{1} + ε_{2})}^{2}}$

3 $\frac{ε_{1} ε_{2}}{ε_{1} + ε_{2}}$

4 $\frac{ε_{0} (ε_{1} + ε_{2})}{2}$

Ans.
(2)

$C_{1} = \frac{ε_{0} A}{\frac{d}{2 ε_{1}} + \frac{d}{2 ε_{2}}} = \frac{ε_{0} A}{d} {\frac{1}{\frac{ε_{2} + ε_{1}}{2 ε_{1} ε_{2}}}}$

$C_{2} = \frac{ε_{0}}{d} (ε_{1} \frac{A}{2} + ε_{2} \frac{A}{2}) = \frac{ε_{0} A}{d} {\frac{ε_{1}}{2} + \frac{ε_{2}}{2}}$

$\frac{C_{1}}{C_{2}} = \frac{2 ε_{1} ε_{2}}{(ε_{2} + ε_{1}) \frac{(ε_{1} + ε_{2})}{2}} = \frac{4 ε_{1} ε_{2}}{{(ε_{1} + ε_{2})}^{2}}$

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