JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A charged particle (mass \(m\) and charge \(q\) ) moves along \(X\) axis with velocity \(V _{0}\). When it passes through the origin it enters a region having uniform electric field \(\overrightarrow{ E }=- E \hat{ j }\) which extends upto \(x = d\). Equation of path of electron in the region \(x > d\) is
- A \(y=\frac{q E d}{m V_{0}^{2}}\left(\frac{d}{2}-x\right)\)
- B \(y=\frac{q E d}{m V_{0}^{2}}(x-d)\)
- C \(y =\frac{ qEd }{ mV _{0}^{2}} x\)
- D \(y =\frac{ qEd ^{2}}{ mV _{0}^{2}} x\)
Answer & Solution
Correct Answer
(A) \(y=\frac{q E d}{m V_{0}^{2}}\left(\frac{d}{2}-x\right)\)
Step-by-step Solution
Detailed explanation
Let particle have charge \(q\) and mass \({ }^{\prime} m ^{\prime}\) Solve for \((q,m)\) mathematically \(F _{ x }=0, a _{ x }=0,( v )_{ x }=\) constant time taken to reach at \(P ^{\prime}=\frac{ d }{ V _{0}}= t _{0}( let )\)\(\ldots(1)\) (Along…
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