The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants , , and (as shown in the figure). If the new capacitance is , then the value of is [

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Solution

Q.
The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_{1} = 2$ , $K_{2} = 3$ , and $K_{3} = 5$ (as shown in the figure). If the new capacitance is $\frac{n}{3} C$ , then the value of $n$ is _______. [2026]

Ans.
(8)

$C_{1} = \frac{5 ε_{0} (A / 2)}{d / 2} = \frac{5 ε_{0} A}{d} = 5 C$

$C_{2} = \frac{2 ε_{0} (A / 2)}{d / 2} = \frac{2 ε_{0} A}{d} = 2 C$

$C_{1}$ and $C_{2}$ in series

$C' = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{(5 C) (2 C)}{7 C} = \frac{10}{7} C$

$C_{3} = \frac{3 ε_{0} (A / 2)}{d / 2} = 3 C$

$C_{4} = \frac{2 ε_{0} (A / 2)}{d / 2} = 2 C$

$C_{4}$ and $C_{3}$ in series; $C'' = \frac{(2 C) (3 C)}{5 C} = \frac{6}{5} C$

$C'$ and $C''$ in parallel;

So, $C_{eq} = C (\frac{6}{5} + \frac{10}{7}) = C (\frac{42 + 50}{35}) = \frac{92}{35} C$

$\frac{92}{35} C = \frac{n C}{3}$

$n = \frac{92 \times 3}{35} = 7.9 \approx 8$

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