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(A) The initial capacitance of the parallel plate capacitor with air as the medium is $C_0 = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickne

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$\frac{d}{2}$

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(A) The initial capacitance of the parallel plate capacitor with air as the medium is $C_0 = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced,the new capacitance $C'$ is given by $C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.Given that the capacity increases by $50 \%$,the new capacitance is $C' = C_0 + 0.5 C_0 = 1.5 C_0 = \frac{3}{2} C_0$.Substituting the expressions,we get $\frac{\epsilon_0 A}{d - t + \frac{t}{K}} = \frac{3}{2} \frac{\epsilon_0 A}{d}$.This simplifies to $\frac{1}{d - t + \frac{t}{3}} = \frac{3}{2d}$.Cross-multiplying gives $2d = 3(d - t + \frac{t}{3}) = 3(d - \frac{2t}{3}) = 3d - 2t$.Rearranging for $t$,we get $2t = 3d - 2d = d$,which implies $t = \frac{d}{2}$.

The plates of a parallel plate capacitor are separated by a distance $d$ with air as the medium between them. $A$ dielectric slab of dielectric constant $K = 3$ is introduced between the plates so as to increase the capacity by $50 \%$. The thickness of the dielectric slab is

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