Q. The area of the plates of a parallel plate condenser is $A$ and the distance between the plates is $10mm.$ There are two dielectric sheets in it, one of dielectric constant 10 and thickness $6mm$ and the other of dielectric constant 5 and thickness $4mm.$ The capacity of the condenser is
A) $\frac{12}{35}\varepsilon_{0}A$
B) $\frac{2}{3}\varepsilon_{0}A$
C) $\frac{5000}{7}\varepsilon_{0}A$
D) $1500\ \varepsilon_{0}A$
Explanation
(c) $C = \frac{\varepsilon_{0}A}{\left( \frac{t_{1}}{k_{1}} + \frac{t_{2}}{k_{2}} \right)} = \frac{\varepsilon_{0}A}{\frac{6 \times 10^{- 3}}{10} + \frac{4 \times 10^{- 3}}{5}} = \frac{5000}{7}\varepsilon_{0}A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.