Q. The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate $= A$)
A) $\frac{\varepsilon_{0}A}{\left( \frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}} + \frac{d_{3}}{K_{3}} \right)}$
B) $\frac{\varepsilon_{0}A}{\left( \frac{d_{1} + d_{2} + d_{3}}{K_{1} + K_{2} + K_{3}} \right)}$
C) $\frac{\varepsilon_{0}A(K_{1}K_{2}K_{3})}{d_{1}d_{2}d_{3}}$
D) $\varepsilon_{0}\left( \frac{AK_{1}}{d_{1}} + \frac{AK_{2}}{d_{2}} + \frac{AK_{3}}{d_{3}} \right)$
Explanation
The correct answer is Option A: $\frac{\varepsilon_{0}A}{\left( \frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}} + \frac{d_{3}}{K_{3}} \right)}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.