Question

1. (18.3.25) Point charges of 5.00 pC and -3.00 uC are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive?

Answer 1

Let a charge of 1 mu c is placed at c, at a dist. X metre from B. Net force of the charge at c F = 1/ 4 pi elementof_0 [5 times 10^-6 times 10^-8/ (0.25 + x)^2 + -3 times 10^-6 times 10^-6/x^2] N when this force is 0 1/ 4 pi elementof 5 times 10^-12/ (0.25 + x)^2 = 1/ 4 pi elementof 3 times 10^-12/ x^2 implies 5/3 = (0.25/x + 1)^2 implies 0.25/x + 1 = squareroot 5/3 = 1.29 implies 0.25/x = 0.29 implies x tildeequal 0.859 m. Let a + 1mu c be placed in between the two charges 5 mu c and 3 mu c at a distance x from 5 mu c Net force F = 1/ 4 pi elementof_0 [5 times 10^-6 times 10^-6/x^2 - 3 times 10^-6 times 10^-6/ (0.25 - x)^2 N [-ve sign is used because the forces will be in the opposite direction] now by