A parallel plate capacitor has capacitance C, | vedclass

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Question

(A) The initial capacitance of the parallel plate capacitor with vacuum is $C = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t = d/3$

Vedclass · Accepted Answer

$\frac{3 KC }{2 K+1}$

Vedclass · Answer

(A) The initial capacitance of the parallel plate capacitor with vacuum is $C = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t = d/3$ and dielectric constant $K$ is introduced, the system can be modeled as two capacitors in series: one with air of thickness $d - t = 2d/3$ and one with dielectric of thickness $t = d/3$.The capacitance of the air part is $C_1 = \frac{\epsilon_0 A}{2d/3} = \frac{3}{2} \frac{\epsilon_0 A}{d} = \frac{3}{2} C$.The capacitance of the dielectric part is $C_2 = \frac{K \epsilon_0 A}{d/3} = 3K \frac{\epsilon_0 A}{d} = 3KC$.Since these are in series, the equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(\frac{3}{2} C) (3KC)}{\frac{3}{2} C + 3KC} = \frac{\frac{9}{2} KC^2}{\frac{3}{2} C (1 + 2K)} = \frac{3KC}{2K + 1}$.

$A$ parallel plate capacitor has capacitance $C$, when there is vacuum within the parallel plates. $A$ sheet having thickness $t = d/3$ (where $d$ is the separation between the plates) and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is:

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