If the force of a charge of 0.2 coulomb in an electric field is 4N, then the electric field intensity of the field is
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Correct Answer: Option C
Explanation:
Field intensity = \(\frac{force}{charge}\)
\(\frac{4}{0.2}\) = 20N/C
Field intensity = \(\frac{force}{charge}\)
\(\frac{4}{0.2}\) = 20N/C