A capacitor has capacitance C_0 when there is | vedclass

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(D) The initial capacitance of the capacitor is $C_0 = \frac{\varepsilon_0 A}{d}$.When two dielectric slabs of thickness $d/2$ are placed between the

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$2 C_0\left(\frac{K_1 K_2}{K_1+K_2}\right)$

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(D) The initial capacitance of the capacitor is $C_0 = \frac{\varepsilon_0 A}{d}$.When two dielectric slabs of thickness $d/2$ are placed between the plates,the arrangement acts as two capacitors in series,each with plate separation $d/2$.The capacitance of the first slab is $C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d} = 2 K_1 C_0$.The capacitance of the second slab is $C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d} = 2 K_2 C_0$.Since they are in series,the equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2 K_1 C_0} + \frac{1}{2 K_2 C_0} = \frac{1}{2 C_0} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{1}{2 C_0} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$.Therefore,$C_{eq} = 2 C_0 \left( \frac{K_1 K_2}{K_1 + K_2} \right)$.

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