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(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{k A \varepsilon_0}{d}$, where $k$ is the dielectric constant, $A$

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$C_2 = 2C_1$

Vedclass · Answer

(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{k A \varepsilon_0}{d}$, where $k$ is the dielectric constant, $A$ is the area of the plates, $\varepsilon_0$ is the permittivity of free space, and $d$ is the separation between the plates.Initially, the capacitance is $C_1 = \frac{k A \varepsilon_0}{d}$.When the plates are moved closer such that the new separation $d' = \frac{d}{2}$, the new capacitance $C_2$ becomes:$C_2 = \frac{k A \varepsilon_0}{d'} = \frac{k A \varepsilon_0}{d/2} = 2 \left( \frac{k A \varepsilon_0}{d} \right)$.Substituting $C_1$ into the equation, we get $C_2 = 2C_1$.

The plates of a parallel plate capacitor of capacity $C_1$ are moved closer together until they are half their original separation. The new capacitance $C_2$ is

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