point particle has a charge equal to +2.30 μC and is fixed at the origin.?
(a) What is the electric potentialV at a point 3.20 m from the origin assuming that V = 0 at infinity?
Update:(a) What is the electric potentialV at a point 3.20 m from the origin assuming that V = 0 at infinity?
(b) How much work must be done to bring a second point particle that has a charge of +2.90 μC from the infinity to a distance of 3.20 m from the +2.30-μC charge?
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Verified answer
A. V = kq/r, where k = 8.99E9 Nm^2/C^2
V = 6461.56 v
B. Potential energy PE = kq1q2/r = 0.018733 J
v=k*q/r
a)
v=9*10^8*(2.3*10^-6)/3.2 = 646.875 V
b)
9*10^8*(2.3*10^-6)*(2.9*10^-6)/3.2 = 0.0018759375 J