Two point charges of + 5x 10-19 C and +10x 10-19 C are separated by a distance of 1 m. At which point on the line joining them is the electric field intensity zero? (UP 2017) Hint: According to the question, 91 q = 5 10-19 C, q2 = 10 10-19 C 92 A P B X -1-x- The electric field at point P will be 0, if: E = E2 Now, substitute the values, we have 1 5 10-19 1 4 10x 10-19 (1-x)2 2x = (1-x) 2x = (1-x) x(1+2)=1 1 x= =041 1+2 i.e., at 0.41 m right of q, electric field due to both charges will be zero.