Electrostatics Ques 115
- 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
$(2000,2 \mathrm{M})$
(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: 115.(d)
Solution:
Formula:
- Applying $C=\frac{\varepsilon_{0} A}{d-t_{1}-t_{2}+\frac{t_{1}}{K_{1}}+\frac{t_{2}}{K_{2}}}$, we have
$\frac{\varepsilon_{0}(A / 2)}{d-d / 2-d / 2+\frac{d / 2}{K_{1}}+\frac{d / 2}{K_{3}}}$
$ +\frac{\varepsilon_{0}(A / 2)}{d-d / 2-d / 2+\frac{d / 2}{K_{2}}+\frac{d / 2}{K_{3}}}=\frac{K \varepsilon_{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}} $
BETA
