Electrostatics

Question: The magnitude of electric field intensity $E$is such that, an electron placed in it would experience an electrical force equal to its weight is given by



1) $mge$
2) $\frac{{mg}}{e}$
3) $\frac{e}{{mg}}$
4) $\frac{{{e^2}}}{{{m^2}}}g$
Solution: Explanation: Explanation
electric-field-potential

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