Electrostatics Question 261
Question: A charged particle q is shot towards another charged particle Q which is fixed, with a speed $ \nu $ . It approaches Q upto a closest distance r and then returns. If q were given a speed $ 2\nu $ , the closest distances of approach would be [AIEEE 2004]
Options:
A) $ 250J $
B) $ 750J $
C) $ 1225J $
D) $ 1475J $
Show Answer
Answer:
Correct Answer: D
Solution:
Energy $ =\frac{1}{2}{\varepsilon _{0}}E^{2}\times (A\times d) $
$ =\frac{1}{2}{\varepsilon _{0}}( \frac{V^{2}}{d^{2}} )Ad $
$ =\frac{1}{2}\times \frac{8.85\times {{10}^{-12}}\times {{(10^{5})}^{2}}\times 25\times 10^{6}}{0.75\times 10^{3}}=1475J $
