A capacitor has capacity C when its parallel | vedclass

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Question

(D) The initial capacitance of the parallel plate capacitor with air as the medium is given by $C = \frac{\epsilon_0 A}{d}$.When a dielectric slab of

Vedclass · Accepted Answer

$\frac{2KC}{K + 1}$

Vedclass · Answer

(D) The initial capacitance of the parallel plate capacitor with air as the medium is given by $C = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t = \frac{d}{2}$ and dielectric constant $K$ is inserted, the new capacitance $C'$ is given by the formula $C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.Substituting the values $t = \frac{d}{2}$ into the formula:$C' = \frac{\epsilon_0 A}{d - \frac{d}{2} + \frac{d}{2K}}$$C' = \frac{\epsilon_0 A}{\frac{d}{2} + \frac{d}{2K}}$$C' = \frac{\epsilon_0 A}{\frac{d}{2} (1 + \frac{1}{K})}$$C' = \frac{2 \epsilon_0 A}{d (\frac{K + 1}{K})}$$C' = \frac{2 \epsilon_0 A}{d} \cdot \frac{K}{K + 1}$Since $C = \frac{\epsilon_0 A}{d}$, we substitute $C$ into the expression:$C' = \frac{2KC}{K + 1}$.

Similar Questions

$A$ capacitor has two circular plates,each with a radius of $8 \ cm$,separated by a distance of $1 \ mm$. Calculate the capacitance of this capacitor when a dielectric slab (dielectric constant $K = 6$) is placed between the plates.

Assertion: When a battery remains connected,the electric potential energy increases if a dielectric material is inserted between the plates of a capacitor.
Reason: When a battery remains connected,the charge on the plates of the capacitor remains the same.

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