Three charges, Q ,- q and 2 q are placed at the vertices of a right-angled isosceles triangle. What is the value of q for the net electrostatic energy of the configuration to be zero?

Step-by-step solution

The electrostatic potential energy U of a system of point charges is given by the sum of the potential energies of all pairs of charges: U = k q_i q_j r_ ij . The charges are Q , 2q , and -q . The distances between the pairs are r , r , and r 2 (hypotenuse of the right-angled isosceles triangle). The total potential energy U is: U = k(Q)(2q) r + k(Q)(-q) r 2 + k(2q)(-q) r = 0 Dividing by kq r (assuming k, q, r 0 ): 2Q - Q 2 - 2q = 0 2q = 2Q - Q 2 2q = Q ( 2 - 1 2 ) q = Q ( 1 - 1 2 2 ) Answer: Q [1- 1 2 2 ]