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(A) The initial capacitance of the parallel plate capacitor with air between the plates is $C_1 = \frac{\varepsilon_0 A}{d}.$When two dielectric slabs

Vedclass · Accepted Answer

$2.66$

Vedclass · Answer

(A) The initial capacitance of the parallel plate capacitor with air between the plates is $C_1 = \frac{\varepsilon_0 A}{d}.$When two dielectric slabs of thickness $t_1 = \frac{3}{8}d$ and $t_2 = \frac{d}{2}$ are inserted,the remaining air gap is $t_{air} = d - (\frac{3}{8}d + \frac{1}{2}d) = d - \frac{7}{8}d = \frac{1}{8}d.$The new capacitance $C_2$ is given by the formula for a capacitor with multiple dielectric slabs:$C_2 = \frac{\varepsilon_0 A}{\frac{t_1}{K_1} + \frac{t_2}{K_2} + t_{air}} = \frac{\varepsilon_0 A}{\frac{3d}{8K_1} + \frac{d}{2K_2} + \frac{d}{8}}.$Given $K_1 = 1.25 K_2 = \frac{5}{4} K_2,$ so $K_2 = \frac{4}{5} K_1 = 0.8 K_1.$Substituting $K_2$ into the expression for $C_2$:$C_2 = \frac{\varepsilon_0 A}{d \left( \frac{3}{8K_1} + \frac{1}{2(0.8K_1)} + \frac{1}{8} \right)} = \frac{\varepsilon_0 A}{d \left( \frac{3}{8K_1} + \frac{1}{1.6K_1} + \frac{1}{8} \right)} = \frac{\varepsilon_0 A}{d \left( \frac{3}{8K_1} + \frac{5}{8K_1} + \frac{1}{8} \right)} = \frac{\varepsilon_0 A}{d \left( \frac{8}{8K_1} + \frac{1}{8} \right)} = \frac{\varepsilon_0 A}{d \left( \frac{1}{K_1} + \frac{1}{8} \right)}.$Given $C_2 = 2 C_1,$ we have:$\frac{\varepsilon_0 A}{d (\frac{1}{K_1} + \frac{1}{8})} = 2 \frac{\varepsilon_0 A}{d} \Rightarrow \frac{1}{\frac{1}{K_1} + \frac{1}{8}} = 2 \Rightarrow \frac{1}{K_1} + \frac{1}{8} = \frac{1}{2}.$$\frac{1}{K_1} = \frac{1}{2} - \frac{1}{8} = \frac{4-1}{8} = \frac{3}{8}.$Therefore,$K_1 = \frac{8}{3} \approx 2.66.$

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