Question

An electron with a kinetic energy of $100\text{eV}$ enters the space between the plates of a parallel plate capacitor made of two dense metal grids at an angle of ${30}^{\circ }$ with the plates of capacitor and leaves this space at an angle of ${45}^{\circ }$ with the plates. What is the potential difference of the capacitor ?

• A. $100\text{V}$
• B. $50\text{V}$
• C. $150\text{V}$
• D. $200\text{V}$

Answer 1

The correct option is B $50\text{V}$

Given, Initial Kinetic Energy $K{E}_i=100\text{eV}$



In between the parallel plates, the electric field $E$ exists perpendicular to the plates. Thus, horizontal component of velocity of the electron remains the same.

$v_1\cos {30}^{\circ }=v_2\cos {45}^{\circ }$

$v_1\frac{\sqrt{3}}{2}=v_2\times \frac{1}{\sqrt{2}}$

$\frac{3v_1^2}{4}=\frac{v_2^2}{2}$

$\frac{3}{2}\times \frac{1}{2}m{v}_1^2=\frac{1}{2}m{v}_2^2$

$\frac{3}{2}\times K{E}_i=K{E}_f$

$\Rightarrow K{E}_f=1.5\times 100=150\text{eV}$

$\therefore \text{Potential difference}=\frac{\Delta KE}{e}=\frac{(150-100)\text{eV}}{e}=50\text{V}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.