Question

The two charges in the figure below are separated by d = 4.00 cm. (Let q1 = ?12.5 nC and q2 = 26.5 nC.) (a) Find the electric potential at point A. kV (b) Find the electric potential at point B, which is halfway between the charges. kV

Answer 1

P(q1) = K q1 / r
where K = Coulomb's constant = 9 * 10^9 ; r = 0.04m

P(q2) = K q2 / r

P(A) = P(q1) + P(q2) = (K/r) ( q1 + q2)

P(A) = 9 * 10^9 [(-12.5 * 10^-9) + (26.5 ^ 10^-9)] / 0.04

P(A) = 9 * 10^9 ( 14* 10^-9) / 0.04

P(A) = 31.5 kV

b) . So the distance between the charges must be 4 cm or 0.04 m.

Therefore, the half way point is 0.04/2 = 0.02m

P(B) = (K/r) (q1 + q2)

where, this time, r = 0.02 m

P(B) = (9 * 10^9 / 0.02) ( 14 * 10^-9) = 6.3kV