Question

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

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

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 $=\frac{e E}{m}$
Now, as $v^{2}=u^{2}+2 a s$
$v^{2}=2 \times \frac{e E}{m} \times s$
$=2 \times\left[1.76 \times 10^{11} \times 10^{4} \times 2 \times 10^{-2}\right]$
$\left[\frac{e}{m}=1.76 \times 10^{11} Ckg ^{-1}\right]$
$=7.04 \times 10^{13}$
$\Rightarrow v=0.85 \times 10^{7}\, ms ^{-1}$