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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........

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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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