Question

An electron is released from the bottom plate $A$ as shown in the figure $\left(E=10^{4}N{C}^{-1}\right)$. The velocity of the electron when it reaches plate $B$ will be nearly equal to

• A. $0.85\times 10^{7}m{s}^{-1}$
• B. $1.0\times 10^{7}m{s}^{-1}$
• C. $1.25\times 10^{7}m{s}^{-1}$
• D. 1. $65\times 10^{7}m{s}^{-1}$

Answer 1

Answer: (A)

Given, $E=10^{4}N{C}^{-1}$,
$s=2\times 10^{-2}m$
$U=0$
The force acting on the electron $=eE$
Acceleration of electron $=\frac{eE}{m}$
Now, as $v^2=u^2+2as$
$v^2=2\times \frac{eE}{m}\times s$
$=2\times \left[1.76\times 10^{11}\times 10^4\times 2\times 10^{-2}\right]$
$\left[\frac{e}{m}=1.76\times 10^{11}Ck{g}^{-1}\right]$
$=7.04\times 10^{13}$
$\Rightarrow v=0.85\times 10^{7}m{s}^{-1}$