A parallel plate capacitor has a capacitance | vedclass

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(A) The initial capacitance of the parallel plate capacitor is $C = \frac{\epsilon_0 A}{d}$.In the second condition,the space between the plates is di

Vedclass · Accepted Answer

$\frac{C}{2}(1 + K)$

Vedclass · Answer

(A) The initial capacitance of the parallel plate capacitor is $C = \frac{\epsilon_0 A}{d}$.In the second condition,the space between the plates is divided into two parts,each with area $A/2$ and distance $d$ between the plates.These two parts act as two capacitors connected in parallel.One capacitor has air as the dielectric,so its capacitance is $C_1 = \frac{\epsilon_0 (A/2)}{d} = \frac{C}{2}$.The other capacitor has a dielectric material with constant $K$,so its capacitance is $C_2 = \frac{K \epsilon_0 (A/2)}{d} = \frac{KC}{2}$.Since they are connected in parallel,the equivalent capacitance is $C_{eq} = C_1 + C_2$.$C_{eq} = \frac{C}{2} + \frac{KC}{2} = \frac{C}{2}(1 + K)$.

$A$ parallel plate capacitor has a capacitance $C$ and the distance between the plates is $d$. If the space between the plates is filled with a dielectric material of dielectric constant $K$ as shown in the figure,find the new capacitance of the capacitor.

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