Question

An electric field of magnitude $1000N{C}^{-1}$ is produced between two parallel plates having a separation of $2.0$cm as shown in figure. Suppose the electron is projected from the lower plate with the speed calculate in part.

Answer 1

$\overrightarrow{E}=1000N/C$
Now, $U=u\cos {60}^o$
$V=0$, $s=?$

Acceleration is given as:

$a=\frac{qE}{m}$

$=\frac{1.6\times {10}^{-19}\times 1000}{9.1\times {10}^{-31}}$

$a=1.75\times {10}^{14}m/s^2$

The displacement of the charge can be given as:

$v^2=u^2-2as$
$\Rightarrow s=\frac{(u\cos {60}^o{)}^2}{2\times a}$

$=\frac{{(2.64\times {10}^6\times \frac{1}{2})}^2}{2\times 1.75\times {10}^{14}}=\frac{1.75\times {10}^{12}}{3.5\times {10}^{14}}$

$=0.497\times {10}^{-2}\approx 0.005m\approx 0.50cm$.