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Grade 12IIT JEE Entrance Exam

Two equally charged small balls placed at a fixed distance experience a force F. A similar uncharged ball, after touching one of them is placed at the middle point between the two balls. The force experienced by this ball is:

Profile image of suyash
10 Years agoGrade 12
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1 Answer

Profile image of Medha
7 Years ago
A----------B the distance b/w them is d. Now C came and touched one ball(suppose A) and then kept at the mid point.
C is uncharged.
So, 0+Q/2 = Q/2 both A and C got charge Q/2.
Further, 
A---------C---------B, distance b/w AC and BC is d/2 respectively. B is applying f2 force. A is applying f1 force.
Charges on them are :
A=Q/2
C=Q/2
B=Q
 
Fnet= f2-f1
        =    4KQ^2/2d^2 - 4KQ^2/4d^2
        =    2KQ^2×d^2 - KQ^2/d^2
    Fnet    =  F

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