Question

An electron is moving with a horizontal velocity of $u=5\times {10}^6\text{m/s}$ before it enters a region of uniform downward electric field. It leaves the region after travelling a horizontal distance of $20\text{cm}$. Electron is deflected by $4\text{cm}$ before it reaches a screen $10\text{cm}$ from the electric field region. If electric field magnitude is $\frac{9100}{16x}\text{N/C}$ , then $x$ is

Answer 1

Just after the exit from the electric field, $\tan \theta =\frac{v_y}{v_x}=\frac{0.04-x}{0.1}$
$\Rightarrow \frac{at}{u}=\frac{0.04-\frac{1}{2}a{t}^2}{0.1}$
$\Rightarrow 0.1at=0.04u-\frac{u}{2}a{t}^2$
$\Rightarrow a(0.1t+0.5u{t}^2)=0.04u$
Now, time $t=\frac{0.2}{u}$
$\therefore a(\frac{0.02}{u}+\frac{0.02}{u})=0.04u$
$\Rightarrow a=u^2$
$\therefore a=(5\times {10}^6{)}^2=25\times {10}^{12}{\text{m/s}}^2$

Now, we know that $a=\frac{eE}{m}$

$\Rightarrow E=\frac{25\times {10}^{12}\times 9.1\times {10}^{-31}}{1.6\times {10}^{-19}}=\frac{9100}{64}$