Questions
Four particles, each of mass m and charge q, are held at the vertices of a square of side 'a', as shown in Fig. 2.132. They are released at t = 0 and move under mutual repulsive forces . Speed of any particle when its.distance from the centre of square doubles, is :
detailed solution
Correct option is A
Loss in potential energy=4kq2a+2kq2a2-4kq2(2a)+2kq2(2a2) =2+12kq2aAs gain in kinetic energy = loss in potential energySo, 4 x 12mv2=2+12kq2a⇒v=1+122q24πε0maTalk to our academic expert!
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