What is the magnitude of the electric field at the location of the test charge?
A positivly test charge of 4.0x10^-6 C is in an electric field that exerts a force of 2.0x10^-4 N on it. What
F = qE
Answer:
E = F/q = 0.000 2 N / 0.000 004 C = 50 V/m
thank you cards
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment