Question

Find the magnitude of uniform electric field $E$ in $N/C$ (direction shown in figure) if an electron entering with velocity $100m/s$ 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

Answer: 100

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