Question

Two point charges 4Q and Q are seperatedby a distance of 1m in air. a) At what point on the line joining the two charges is the electric field intensity zero? b) Also calculate the electrostatic potential energy of the system of two charges taking Q= 2x10 -7 C.

Answer 1

let electric field zero from 4Q be x
then
1/4π€° 4Q/x² = 1/4π€• Q/(1-x)²
4(1-x)² = x²
2(1-x) = x
x = 2/3m
or x = 66.67 cm from 4Q charge
(ii)
potential energy = (9×10power9) × 4( 2x10 -7 )²/1²
= 144 × 10 -5 joule