In the figure below, point P is at distance d1 = 3.00 m from particle 1 (q1 = -2
ID: 2173607 • Letter: I
Question
In the figure below, point P is at distance d1 = 3.00 m from particle 1 (q1 = -2e) and distance d2 = 1.50 m from particle 2 (q2 = +1e), with both particles fixed in place.
(a) With V = 0 at infinity, what is V at P?
(b) If we bring a particle of charge q3 = +2e from infinity to point P, how much work do we do?
(c) What is the potential energy of the three-particle system once the third particle is in place?
I have found that the answers for (a) and (b) were both 0. The answers are correct, although I am not entirely sure if I used the proper calculations to find them, since now I am having problems with figuring out (c). Any and all help would be appreciated.
Explanation / Answer
a.v1 = q1/4r1 = 9*109 *-2*e/3 = -6*109*e V
v2 = q2/4r2 = 9*109 *e/1.5 = 6*109*e V
thus total potential at p = 6*109*e V-6*109*e V = 0V
b. since potential at P is 0V, so work done to bring q3 = 0J