(A) The original capacitance of the parallel plate capacitor is given by $C_1 = \frac{\epsilon_0 A}{d}$.When a metal sheet of thickness $t = \frac{d}{

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(A) The original capacitance of the parallel plate capacitor is given by $C_1 = \frac{\epsilon_0 A}{d}$.When a metal sheet of thickness $t = \frac{d}{2}$ is introduced between the plates,the effective distance between the plates decreases.The new capacitance $C_2$ is given by the formula $C_2 = \frac{\epsilon_0 A}{d - t}$.Substituting $t = \frac{d}{2}$,we get $C_2 = \frac{\epsilon_0 A}{d - \frac{d}{2}} = \frac{\epsilon_0 A}{d/2} = \frac{2 \epsilon_0 A}{d}$.Therefore,the ratio of the new capacitance to the original capacitance is $\frac{C_2}{C_1} = \frac{2 \epsilon_0 A / d}{\epsilon_0 A / d} = \frac{2}{1}$ or $2:1$.

Class 12 Physics Electric Potential and Capacitance Effect of Dielectric Inside Capacitor

Medium

Two metallic plates form a parallel plate capacitor. The distance between the plates is $d$. $A$ metal sheet of thickness $\frac{d}{2}$ and of area equal to the area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

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A
$2: 1$
B
$1: 2$
C
$1: 4$
D
$4: 1$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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