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QJEEM23Ph0b2f: JEE Main 2023 Physics - Class 12 JEEM23Ph0b2f Solution: A uniform electric field of 10 N/C is created between two…

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 ( θ\theta ) of deviation of the path of electron as it comes out of the field is ___________ (in degree).

JEE Main 2023 (Online) 25th January Morning Shift Physics - Electrostatics Question 70 English

Step-by-Step Verified Solution

amp;0.5e=12mvx2⇒vx=emamp; Along xL=vxt=emtamp; Along y vy=eEmtamp;dividing⁡vyL=Eem=Evxamp;⇒tan⁡θ=vyvx=E×L=10×0.1=1amp;θ=45∘\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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