A charge of 1 x \(10^{-5}\)C experiences a force of 40N at a certain point in space. What is the magnitude of the electric field intensity at the point in Newton and couloumb
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Correct Answer: Option B
Explanation:
\(E= \frac{F}{O}= \frac{40}{1\times10^{-5}}=4 \times10^{6}\)
\(E= \frac{F}{O}= \frac{40}{1\times10^{-5}}=4 \times10^{6}\)