A parallel plate capacitor with plate area A | vedclass

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(C) The figure shows a series combination of two capacitors with capacitances $C_{1}$ and $C_{2}$,where the plates are separated by a distance $d/2$ e

Vedclass · Accepted Answer

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

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(C) The figure shows a series combination of two capacitors with capacitances $C_{1}$ and $C_{2}$,where the plates are separated by a distance $d/2$ each.The individual capacitances are given by:$C_{1} = \frac{K_{1} \varepsilon_{0} A}{d / 2} = \frac{2 K_{1} \varepsilon_{0} A}{d}$$C_{2} = \frac{K_{2} \varepsilon_{0} A}{d / 2} = \frac{2 K_{2} \varepsilon_{0} A}{d}$Since the capacitors are in series,the equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$$C_{eq} = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$Substituting the values:$C_{eq} = \frac{(\frac{2 K_{1} \varepsilon_{0} A}{d}) (\frac{2 K_{2} \varepsilon_{0} A}{d})}{\frac{2 K_{1} \varepsilon_{0} A}{d} + \frac{2 K_{2} \varepsilon_{0} A}{d}}$$C_{eq} = \frac{4 K_{1} K_{2} \varepsilon_{0}^2 A^2 / d^2}{\frac{2 \varepsilon_{0} A}{d} (K_{1} + K_{2})}$$C_{eq} = \frac{2 \varepsilon_{0} A}{d} \left( \frac{K_{1} K_{2}}{K_{1} + K_{2}} \right)$

$A$ parallel plate capacitor 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

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