Two particles A and B , possessing charges of +2.00 10⁻⁶ C and -4.00 10⁻⁶ C respectively, are held fixed at a separation of 20.0 cm . Find the point(s) on the line AB at which the electric potential is zero.

Step-by-step solution

Let A be at the origin and B be at a distance of 20 cm from A . The electric potential at a point P is given by V = 1 4 ₀ ( q_A r_A + q_B r_B ) . For V = 0 , we have q_A r_A = - q_B r_B . Substituting the given charges: 2.00 10⁻⁶ r_A = 4.00 10⁻⁶ r_B r_B = 2r_A Case 1: P lies between A and B . r_A + r_B = 20 r_A + 2r_A = 20 3r_A = 20 r_A = 20 3 cm This point is 20 3 cm from A along AB . Case 2: P lies outside the segment AB . Since |q_A| r_B - r_A = 20 2r_A - r_A = 20 r_A = 20 cm This point is 20 cm from A along BA