JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 26)
A uniform electric field of 10 N/C 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 eV. The length of each plate is 10 cm. The angle ($$\theta$$) of deviation of the path of electron as it comes out of the field is ___________ (in degree).
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Answer
45
Explanation
$$
\begin{aligned}
& 0.5 \mathrm{e}=\frac{1}{2} \mathrm{mv}_{\mathrm{x}}^2 \Rightarrow \mathrm{v}_{\mathrm{x}}=\sqrt{\frac{\mathrm{e}}{\mathrm{m}}} \\\\
& \text { Along } \mathrm{x} \mathrm{L}=\mathrm{v}_{\mathrm{x}} \mathrm{t}=\sqrt{\frac{\mathrm{e}}{\mathrm{m}}} \mathrm{t} \\\\
& \text { Along y } \mathrm{v}_{\mathrm{y}}=\frac{\mathrm{eE}}{\mathrm{m}} \mathrm{t} \\\\
& \operatorname{dividing} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{L}}=\mathrm{E} \sqrt{\frac{\mathrm{e}}{\mathrm{m}}}=\mathrm{E} v_{\mathrm{x}} \\\\
& \Rightarrow \operatorname{tan} \theta=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\mathrm{E} \times \mathrm{L}=10 \times 0.1=1 \\\\
& \theta=45^{\circ}
\end{aligned}
$$
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