Electrostatics Question 250

Question: An elementary particle of mass $ m $ and charge $ +e $ is projected with velocity $ v $ at a much more massive particle of charge $ Ze, $ where $ Z>0. $ What is the closest possible approach of the incident particle [Orissa JEE 2002]

Options:

A) $ r^{2} $

B) r

C) $ \frac{1}{r} $

D) $ \frac{1}{r^{2}} $

Show Answer

Answer:

Correct Answer: B

Solution:

$ E _{x}=-\frac{dV}{dx}=-ky $ ;

$ E _{y}=-\frac{dV}{dy}=-kx $

therefore $ E=\sqrt{E _{x}^{2}+E _{y}^{2}}=k\sqrt{x^{2}+y^{2}}=kr $

therefore E µ r



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