JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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........
- A \(\tan ^{-1} (4)\)
- B \(\tan ^{-1}(2)\)
- C \(\tan ^{-1}\left(\frac{1}{3}\right)\)
- D \(\tan ^{-1} (3)\)
Answer & Solution
Correct Answer
(B) \(\tan ^{-1}(2)\)
Step-by-step Solution
Detailed explanation
\(a _{ y }=\frac{ F _{ y }}{ m }=\frac{ e ( E )}{ m }=\frac{ e \left(\frac{8\,m }{ e }\right)}{ m }=8\,m / s ^{2}\) \(s _{ x }= u _{ x } t\) \(1=2 \times t\) \(t =\frac{1}{2} sec\) \(v _{ y }= u _{ y }+ a _{ y } t\) \(v _{ y }=0+8 \times \frac{1}{2}\) \(v _{ y }=4\,m / s\)…
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