JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2022
- Hard
- MCQ
A uniform electric field \(E =(8\,m / e ) V / m\) is created between two parallel plates of length \(1 m\) as shown in figure, (where \(m =\) mass of electron and \(e=\) charge of electron). An electron enters the field symmetrically between the plates with a speed of \(2\,m / s\). The angle of the deviation \((\theta)\) of the path of the electron as it comes out of the field will be........

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\tan ^{-1}(2)\)
Step-by-step Solution
Detailed explanation
\(a _{ y }=\frac{ F _{ y }}{ m }=\frac{ e ( E )}{ m }=\frac{ e \left(\frac{8\,m }{ e }\right)}{ m }=8\,m / s ^{2}\)
\(s _{ x }= u _{ x } t\)
\(1=2 \times t\)
\(t =\frac{1}{2} sec\)
\(v _{ y }= u _{ y }+ a _{ y } t\)
\(v _{ y }=0+8 \times \frac{1}{2}\)
\(v _{ y }=4\,m / s\)
\(\tan \theta=\frac{ v _{ y }}{ v _{ x }}=\frac{4}{2}=2 \Rightarrow \theta=\tan ^{-1}(2)\)
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