Question

Two insulating plates are uniformly charged in such a way that potential difference between them is $V_2-V_1=20\text{V}$. Plates are separated by $d=0.1\text{m}$. An electron is released from rest in between the plates from the surface of plate $1$. What is its speed when it hits the plate $2$?
$($Charge on an electron, $e=-1.6\times {10}^{-19}\text{C}$ and mass of an electron, $m_e=9.1\times {10}^{-31}\text{kg})$

• A. $1.87\times {10}^6\text{m/s}$
• B. $32\times {10}^{-19}\text{m/s}$
  
• C. $7.02\times {10}^{12}\text{m/s}$
• D. $2.65\times {10}^6\text{m/s}$

Answer 1

The correct option is D $2.65\times {10}^6\text{m/s}$

Given that,
Potential difference between plates, $\Delta V=V_2-V_1=20\text{V}$
Distance between the plates, $d=0.1\text{m}$
Charge of an electron, $e=-1.6\times {10}^{-19}\text{C}$
Mass of an electron, $m_e=9.1\times {10}^{-31}\text{kg}$

Let $v$ is the velocity of the electron when it hits plate $2$.

By law of conservation of mechanical energy,
Gain in kinetic energy $=$ Loss in potential energy.
$\Rightarrow \frac{1}{2}m{v}^2=e(V_2-V_1)$

$\Rightarrow v=\sqrt{\frac{2e(V_2-V_1)}{m}}$
$\Rightarrow v=\sqrt{\frac{2\times 1.6\times {10}^{-19}\times 20}{9.1\times {10}^{-31}}}$
$\Rightarrow v=2.65\times {10}^6\text{m/s}$

Hence, option (b) is correct.