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Question

Physics Question on Electric Field

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

A

mgem\,g\,e

B

mge\frac{mg}{e}

C

emg\frac{e}{mg}

D

e2m2g\frac{e^2}{m^2} g

Answer

mge\frac{mg}{e}

Explanation

Solution

Force on electron
F=qE=eE=mg| F |= qE = eE = mg
E=mgeE =\frac{ mg }{ e }