A parallel plate condenser with plate area A and separation d is filled with two dielectric material

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1. Electric Charges and Fields

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A parallel plate condenser with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the figure. The dielectric constants are $K_1$ and $K_2$ respectively. The capacitance will be

$\frac{{{\varepsilon _0}A}}{d}\left( {{K_1} + {K_2}} \right)$

$\frac{{{\varepsilon _0}A}}{d}\left( {\frac{{{K_1} + {K_2}}}{{{K_1}{K_2}}}} \right)$

$\frac{{2{\varepsilon _0}A}}{d}\left( {\frac{{{K_1} {K_2}}}{{{K_1}+{K_2}}}} \right)$

$\frac{{2{\varepsilon _0}A}}{d}\left( {\frac{{{K_1} + {K_2}}}{{{K_1}{K_2}}}} \right)$

Solution

The figure shows series combination two capacitors of capacitance $\mathrm{C}_{1}$ and $\mathrm{C}_{2},$ where

$\mathrm{C}_{1}=\frac{\mathrm{K}_{1} \varepsilon_{0} \mathrm{A}}{\mathrm{d} / 2} \quad$ and $\quad \mathrm{C}_{2}=\frac{\mathrm{K}_{2} \varepsilon_{0} \mathrm{A}}{\mathrm{d} / 2}$

$\mathrm{C}^{\prime}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}=\frac{2 \varepsilon_{0} \mathrm{A}}{\mathrm{d}}\left(\frac{\mathrm{K}_{1} \mathrm{K}_{2}}{\mathrm{K}_{1}+\mathrm{K}_{2}}\right)$

Standard 12

Physics

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normal

A parallel plate condenser with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the figure. The dielectric constants are $K_1$ and $K_2$ respectively. The capacitance will be

Solution

Similar Questions

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