WBJEE · Physics · Work Power Energy
A particle of mass \(M\) and charge \(q\), initially at rest, is accelerated by a uniform electric field \(E\) through a distance \(D\) and is then allowed to approach a fixed static charge \(Q\) of the same sign. The distance of the closest approach of the charge \(q\) will then be
- A \(\frac{q Q}{4 \pi \varepsilon_{0} D}\)
- B \(\frac{Q}{4 \pi \varepsilon_{0} E D}\)
- C \(\frac{q Q}{2 \pi \varepsilon_{0} D^{2}}\)
- D \(\frac{Q}{4 \pi \varepsilon_{0} E}\)
Answer & Solution
Correct Answer
(B) \(\frac{Q}{4 \pi \varepsilon_{0} E D}\)
Step-by-step Solution
Detailed explanation
At the distance of closest approach, the entire \(\mathrm{KE}\) of a particle is converted into electric potential energy. i.e.. \(\mathrm{KE}=\mathrm{PE}\) Here, KE of the particle = work done in moving the particle in uniform electric field (E) through a distance (D)…
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