Electrostatics

Question: The electric field due to a charge at a distance of 3 m from it is 500 N/coulomb. The magnitude of the charge is $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\frac{{N - {m^2}}}{{coulom{b^2}}}} \right]$



1) 2.5 micro-coulomb
2) 2.0 micro-coulomb
3) 1.0 micro-coulomb
4) 0.5 micro-coulomb
Solution: Explanation: Explanation
electric-field-potential

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