ElectrostaticsHard
Question
A charged particle q is shot towards another charged particle Q which is fixed, with a speed v it approaches Q upto a closest distance r and then returns. If q were given a speed 2v, the closest distances of approach would be
Options
A.r
B.2r
C.r/2
D.r/4
Solution
By principle of conservation of energy
mv2 = 
....(i)
Finally, m(2v)2 = 
....(ii)
Equation (i) - (ii),
⇒ r′ =
.
mv2 = ....(i)Finally,
m(2v)2 = ....(ii) Equation (i) - (ii),
⇒ r′ =
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