First slide

Capacitance

Question

A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants $K_{1}, K_{2}, K_{3} a n d K_{4}$ as shown in the figure.
If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

Difficult

A

$k = k_{1} + k_{2} + k_{3} + 3 k_{4}$
B

$k = \frac{2}{3} (k_{1} + k_{2} + k_{3}) + 2 k_{4}$
C

$\frac{2}{k} = \frac{3}{k_{1} + k_{2} + k_{3}} + \frac{1}{k_{4}}$
D

$\frac{1}{k} = \frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{k_{3}} + \frac{3}{2 k_{4}}$

Solution

Here,
$\begin{array}{l} C_{1} = \frac{2 ε_{0} k_{1} A}{3 d}, C_{2} = \frac{2 ε_{0} k_{2} A}{3 d} \\ C_{3} = \frac{2 ε_{0} k_{3} A}{3 d}, C_{4} = \frac{2 ε_{0} k_{4} A}{d} \end{array}$
$Given system of C_{1}, C_{2}, C_{3} and C_{4} can be simplified as$
$∴ \frac{1}{C_{A B}} = \frac{1}{C_{1} + C_{2} + C_{3}} + \frac{1}{C_{4}}$
$Suppose, C_{A B} = \frac{k ε_{0} A}{d}$
$\frac{1}{k (\frac{ε_{0} A}{d})} = \frac{1}{\frac{2}{3} \frac{ε_{0} A}{d} (k_{1} + k_{2} + k_{3})} + \frac{1}{\frac{2 ε_{0} A}{d} k_{4}}$
$\begin{array}{l} \Rightarrow \frac{1}{k} = \frac{3}{2 (k_{1} + k_{2} + k_{3})} + \frac{1}{2 k_{4}} \\ ∴ \frac{2}{k} = \frac{3}{k_{1} + k_{2} + k_{3}} + \frac{1}{k_{4}} \end{array}$

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