Question

Find the magnitude of uniform electric field $E$ in $N / C$ (direction shown in figure) if an electron entering with velocity $100 \,m / s$ and making $30^{\circ}$ with the $x$-direction comes out making $60^{\circ}$ with the $x$-direction, after a time numerically equal to $\frac{m}{e}$ of electron.

Answer 1

Applying $v=u+a t$ in $y$ direction, we have
$100 \cos 30^{\circ} \tan 60^{\circ}=100 \sin 30^{\circ}+\frac{e E}{m} t$
( $\because$ At entry point and exit point, $v_{x}=100 \cos 30^{\circ}$
$\therefore $ At exit, $\frac{v_{y}}{100 \cos 30^{\circ}}=\tan 60^{\circ}$
$\left.\Rightarrow v_{y}=100 \cos 30^{\circ} \times \tan 60^{\circ}\right) $
$\therefore \frac{e E}{m} t=100 \times \frac{\sqrt{3}}{2} \cdot \sqrt{3}-100 \times \frac{1}{2}=100$
$\Rightarrow \frac{e}{m} E \cdot \frac{m}{e}=100 $
$\Rightarrow E=100 \,N / C$