Figure shows a ball having a charge q fixed at a point A. Two identical balls having charges +q and –q and mass ‘m’ each are attached to the ends of a light rod of length 2a. The rod is free to rotate about a fixed axis perpendicular to the plane of the paper and passing through the mid-point of the rod. The system is released from the situation shown in figure. Find the angular velocity of the rod when the rod becomes horizontal.


1. 2q3π0 ma3                         

2. q3π0 ma3

3. q6π0 ma3                       

4. 2q4π0 ma3

Concept Videos :-

#1 | Defining Electric Potential Energy
#10 | Electric Potential Energy

Concept Questions :-

Electric potential energy

3. Initial potential energy of the system = 0    ..…(i)

    The rod will rotate in clockwise sense.

    When rod becomes horizontal

      Uf=-Kq2a+Kq23a=-2Kq23a                   ......(ii)

     loss in P.E. = 2Kq23a = gain in K.E.

       2Kq23a=2×12×ma2ω2     using 122        ω2=q26πε0ma3         ω=q6πε0ma3

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