Magnitude of electric field intensity E is such that, an electron placed in it would experience an e

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1. Electric Charges and Fields

easy

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

A

$mge$

B

$\frac{{mg}}{e}$

C

$\frac{e}{{mg}}$

D

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

Solution

(b)According to the question, $eE = mg$ $==>$ $E = \frac{{mg}}{e}$

Standard 12

Physics

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(AIIMS-2008)