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(D) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ is given by $C = \frac{\epsilon_0 A}{d - t + (t/K)}$.For a m

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increases

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(D) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ is given by $C = \frac{\epsilon_0 A}{d - t + (t/K)}$.For a metal sheet (aluminum),the dielectric constant $K$ is infinite $(K = \infty)$.Given that the thickness $t$ is negligible $(t \approx 0)$,we analyze the effect.However,if we place a conducting sheet of thickness $t$ inside,the effective distance between the plates becomes $(d - t)$.The new capacitance is $C' = \frac{\epsilon_0 A}{d - t}$.Since $(d - t) Even if the thickness is negligible,the presence of a conducting plate effectively reduces the separation distance between the plates,leading to an increase in capacitance.

$A$ thin aluminum sheet of negligible thickness is placed between the plates of a parallel plate capacitor. The capacitance of the capacitor:

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