a force of 2.0 x 10^-4 N on it. What is the magnitude of the electric field at the location of the test charge?
A 5.0 X 10^-1 N/C
B 2.0 X 10^-2 N/C
C 4.0 X 10^-2 N/C
D 8.0 X 10^-2 N/C
A positive test charge of 4.0 x 10^-6 C is in an electric field that exerts a force of 2.0 x 10^-4 N on it. W
Answer: A (5.0 X 10^-1 N/C)
Reply:F = Q*E
F is the force.
Q the charge of the test charge
E is the field strength at the location of the test charge
E = F/Q = 2.0*10^-4/4.0*10^-6 = 50 N/C
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