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Answer to Question #3835 in Electromagnetism for mia

Question #3835
A positive test charge of 5.0 × 10–4 C is in an electric field that exerts a force of 2.5 × 10–4 N on it. What is the magnitude of the electric field at the location of the test charge?
Expert's answer
We have that F=Eq, whence
E = F / q = 2.5 × 10–4 N / 5.0 × 10–4 C = 0.5 N/C

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