JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A positive charge particle of \(100 \,mg\) is thrown in opposite direction to a uniform electric field of strength \(1 \times 10^{5} \,NC ^{-1}\). If the charge on the particle is \(40 \,\mu C\) and the initial velocity is \(200 \,ms ^{-1}\), how much distance (in \(m\)) it will travel before coming to the rest momentarily
- A \(1\)
- B \(5\)
- C \(10\)
- D \(0.5\)
Answer & Solution
Correct Answer
(D) \(0.5\)
Step-by-step Solution
Detailed explanation
Distance travelled by particle before stopping \(\frac{ V ^{2}}{2 a }= S \Rightarrow \frac{ v ^{2} m }{2 qE }\) \(\Rightarrow \frac{(200)^{2} \times 100 \times 10^{-6}}{2 \times 40 \times 10^{-6} \times 10^{5}}=0.5 \,m\)
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