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
(Mass of $e^{–}=m$)

• A. $mge$
• B. $\frac{mg}{e}$
• C. $\frac{e}{mg}$
• D. $\frac{e^2}{m^2}g$

Answer 1

The correct option is B $\frac{mg}{e}$

Given:
mass of electron $=m$
charge on each electron $=e$
and acceleration due to gravity$=g$

According to question,

$\text{weight of electron}=\text{electrical force experienced by electron}$

$\Rightarrow \text{mass}\times \text{acceleration due to gravity}$

$=\text{charge}\times \text{electric field intensity}$

$mg=eE$

$E=\frac{mg}{e}$