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
