Q 12-01-076JEE MainJEE Main 2023 (25 Jan, Shift 1)Medium
A uniform electric field of $10\ \text{N C}^{-1}$ is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy $0.5\ \text{eV}$. The length of each plate is $10\ \text{cm}$. The angle ($\theta$) of deviation of the path of electron as it comes out of the field is ______ (in degree).
Numerical value type. Enter your answer.
Answer: 45
Time in the field $t=\dfrac Lv$; transverse velocity gained $v_y=\dfrac{eE}{m}t$.
$$\tan\theta=\frac{v_y}{v}=\frac{eEL}{mv^2}=\frac{eEL}{2K}$$
With $K=0.5\ \text{eV}$: $\tan\theta=\dfrac{10\times0.1}{2\times0.5}=1$, so $\theta=45^\circ$.
Solution by Sreeraj P, M.Sc Physics