Q.
In an isolated parallel plate capacitor of capacitance $C,$ the four surface have charges $Q_{1},Q_{2},Q_{3}$ and $Q_{4}$ as shown. The potential difference between the plates is
A) $\frac{Q_{1} + Q_{2} + Q_{3} + Q_{4}}{2C}$
B) $\frac{Q_{2} + Q_{3}}{2C\ }$
C) $\frac{Q_{2} - Q_{3}}{2C}$
D) $\frac{Q_{1} + Q_{4}}{2C}$
Explanation
(c) Plane conducting surfaces facing each other must have equal and opposite charge densities. Here as the plate areas are equal, $Q_{2} = - Q_{3}$.
The charge on a capacitor means the charge on the inner surface of the positive plate [Here it is $Q_{2}$]
Potential difference between the plates
$= \frac{charge}{capacitance} = \frac{Q_{2}}{C} = \frac{2Q_{2}}{2C}$
$= \frac{Q_{2} - ( - Q_{2})}{2C} = \frac{Q_{2} - Q_{3}}{2C}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.