SATHEE: Electrostatics 5 Question 28

Electrostatics 5 Question 28

30. A parallel plate capacitor of area $A$ , plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constants $K_{1}, K_{2}$ and $K_{3}$ as shown. 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

$(2000, 2 M)$

(a) $\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{2 K_{3}}$

(b) $\frac{1}{K} = \frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}$

(c) $\frac{1}{K} = \frac{K_{1} K_{2}}{K_{1} + K_{2}} + 2 K_{3}$

(d) $K = \frac{K_{1} K_{3}}{K_{1} + K_{3}} + \frac{K_{2} K_{3}}{K_{2} + K_{3}}$

Show Answer

Answer:

Correct Answer: 30. (d)

Solution:

Applying $C = \frac{ε_{0} A}{d - t_{1} - t_{2} + \frac{t_{1}}{K_{1}} + \frac{t_{2}}{K_{2}}}$ , we have

$\frac{ε_{0} (A / 2)}{d - d / 2 - d / 2 + \frac{d / 2}{K_{1}} + \frac{d / 2}{K_{3}}}$

$+ \frac{ε_{0} (A / 2)}{d - d / 2 - d / 2 + \frac{d / 2}{K_{2}} + \frac{d / 2}{K_{3}}} = \frac{K ε_{0} A}{d}$

Solving this equation, we get

$K = \frac{K_{1} K_{3}}{K_{1} + K_{3}} + \frac{K_{2} K_{3}}{K_{2} + K_{3}}$

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