Q. The potentials of the two plates of capacitor are $+ 10V$ and $- 10\ V.$ The charge on one of the plates is $40\ C.$ The capacitance of the capacitor is
A) $2\ F$
B) $4\ F$
C) $0.5\ F$
D) $0.25\ F$
Explanation
(a) The potential difference across the parallel plate capacitor is $10V - ( - 10V) = 20V$
Capacitance $= \frac{Q}{V} = \frac{40}{20} = 2F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.