The magnitude of electric field intensity E s | vedclass

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(B) The force experienced by an electron of charge $e$ in an electric field $E$ is given by $F_e = eE$.According to the problem,this electrical force

Vedclass · Accepted Answer

$\frac{mg}{e}$

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(B) The force experienced by an electron of charge $e$ in an electric field $E$ is given by $F_e = eE$.According to the problem,this electrical force must be equal to the weight of the electron,which is $F_g = mg$,where $m$ is the mass of the electron and $g$ is the acceleration due to gravity.Setting these two forces equal: $eE = mg$.Solving for the electric field intensity $E$,we get: $E = \frac{mg}{e}$.Therefore,the correct option is $B$.

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

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