Find the potential difference between Q and -Q at a distance of 4 m and the velocity of a point charge from point A to B.

Question

Two charges, Q and -Q, are held stationary at a distance of 4 m. Q= 5x10^-6 is given. a) What is the potential difference VB-VA? b) A point charge of mass 0.3 g and charge 2x10^-6 C, initially at rest, leaves point A. What is the velocity at B?

(The distance between Q and point A is 2 m. The distance between A and B are 1 m. The distance between B and -Q is 1 m.)

Answer 1

V_A = k_eQ/2m + (-k_eQ/2m = 0 V_B = -k_eQ/m + k_eQ/3m

and take the difference as V is a scalar quantity.

1/2 mv² = -qV_{A to B} or the gain in kinetic energy iis equal to the loss in electric potential energy. Remember to use m in kg. (The positive charge is going "downhill" in terms of the voltage)