What is the force experienced by a test charge of 0.20 µC placed in an electric field of 3.2 × 106 N/C?

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1 Answers

Given: q0 = 0.20 µC = 0.2 × 106 C,

E = 3.2 × 106 N/C

To find: Force (F)

Formula: E = \(\frac{F}{q_0}\)

Calculation: From formula,

F = Eq0

∴ F = 3.2 × 106 × 0.2 × 10-6 = 0.64 N

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