JEE Main & Advanced
Physics
Electrostatics & Capacitance
Question Bank
Electric Field and Potential
question_answer
A charge of 10 e.s.u. is placed at a distance of 2 cm from a charge of 40 e.s.u. and 4 cm from another charge of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs) [CPMT 1976; MP PET 1989]
A) 87.5
B) 112.5
C) 150
D) 250
Correct Answer:
D
Solution :
Energy \[=\frac{10\times 40}{2}+\frac{10\times 20}{4}=250\,\,erg\]