JEE Physics 2023 Question with Solution

Given: A uniform electric field of $10 \, \text{N/C}$ acts between parallel plates. The electron enters with kinetic energy $5 \, \text{eV}$. The plate length is $10 \, \text{cm}$.

Find: The angle of deviation $\theta$ as the electron emerges from the field.

From the given solution, the kinetic energy of the electron is

$$\frac{1}{2}mv^2 = 5 \, \text{eV}$$

Convert $5 \, \text{eV}$ to joules:

$$5 \times 1.6 \times 10^{-19} = 8 \times 10^{-19} \, \text{J}$$

So,

$$v_x = \sqrt{\frac{2 \times 8 \times 10^{-19}}{9.1 \times 10^{-31}}}$$

The solution then uses the relation

$$\tan \theta = \frac{E_y}{E_x}$$

Substituting the values given in the working:

$$\theta = \tan^{-1}(1) = 45^\circ$$

Therefore, the angle of deviation is $45^\circ$.

Given: The electron moves through a uniform electric field between parallel plates.

Find: The deviation angle of its path at the exit.

The extracted solution states that the deviation depends on the balance between horizontal and vertical effects on the electron. It first computes the electron speed from kinetic energy and then concludes that the ratio used for direction gives

$$\tan^{-1}(1)$$

Hence,

$$\theta = 45^\circ$$

So the required numerical value is 45.