# What is the net force exerted by the two 2.20-µC charges on the test charge q?

Given two particles with Q = 2.20-µC charges as shown in the figure below and a particle with charge q = 1.39 10-18 C at the origin. (Note: Assume a reference level of potential V = 0 at r = .)

(a) What is the net force exerted by the two 2.20-µC charges on the test charge q?
N

(b) What is the electric field at the origin due to the two 2.20-µC particles?
N/C

(c) What is the electrical potential at the origin due to the two 2.20-µC particles?
kV
asked Jan 20, 2012 in Physics

+1 vote

Part A and B are both equal to zero because the two are vector quantities, and because both have equal charges the cancel each other out.

Part C can be found using the equation for electric potential which is:

V = ke * q/r

ke = 8.988*10^9 N*m^2/C^2

q = 2.2 *10^-6 C

r = 0.8 m

plug in and solve for V, which turns out to be  24695 V

since there are two charges just multiply V by 2 to get the total eletric potential at the point.

2*V = 49390 V = 49.4 kV
answered Jan 26, 2012 by anonymous
selected Jan 26, 2012 by awesome

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