A charged particle q is shot towards another charged particle Q which is fixed, with a speed 
. It approaches Q upto a closest distance r and then returns. If q were given a speed 
, the closest distances of approach would be
Text Solution
Verified by ExpertsThe correct answer is:
D
Charge q will momentarily come to rest at a distance r from charge Q when all it's kinetic energy converted to potential energy i.e. 
Therefore the distance of closest approach is given by
⇒ ⇒ 
Hence if v is doubled, r becomes one fourth.
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