JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2023
- Hard
- Numerical
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).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(45\)
Step-by-step Solution
Detailed explanation
\(0.5\,e =\frac{1}{2}\,mv _{ x }^2 \Rightarrow v _{ x }=\sqrt{\frac{ e }{ m }}\)
Along \(x L =v_x t=\sqrt{\frac{ e }{ m }} t\)
Along y \(v _{ y }=\frac{e e }{ m } t\)
\(\operatorname{dividing} \frac{ v _{ y }}{ L }= E \sqrt{\frac{ e }{ m }}= Ev _{ x }\)
\(\Rightarrow \operatorname{Tan} \theta=\frac{ v _{ y }}{ v _{ x }}= E \times L =10 \times 0.1=1\)
\(\theta=45^{\circ}\)
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