Question

A uniform electric field E = (8m/e) V/m is created between two parallel plates of length 1 m as shown in figure, (where m = mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2 m/s. The angle of the deviation ($\theta$) of the path of the electron as it comes out of the field will be ____. [JEE (Main) - 2022]

• A. tan$^{-1}$ (3)
• B. tan$^{-1}$ (2)
• C. tan$^{-1}$ (1/3)
• D. tan$^{-1}$ (4)

Answer 1

Answer: (B)

tan$^{-1}$ (2)

Answer 2

1. Calculate the upward force on the electron due to the downward electric field: $F = eE = e(8m/e) = 8m$.
2. Calculate the upward acceleration of the electron: $a_y = F/m = 8m/m = 8 \text{ m/s}^2$. The horizontal acceleration is $a_x = 0$.
3. The time taken to cross the plates horizontally is $t = \text{length}/\text{horizontal velocity} = 1/2 = 0.5 \text{ s}$.
4. Calculate the final vertical velocity: $v_y = \text{initial vertical velocity} + a_y t = 0 + 8(0.5) = 4 \text{ m/s}$.
5. The final horizontal velocity is the same as the initial horizontal velocity: $v_x = 2 \text{ m/s}$.
6. The angle of deviation $\theta$ is given by $\tan \theta = v_y/v_x = 4/2 = 2$.
7. Therefore, $\theta = \tan^{-1}(2)$.