A charged particle of charge Q is held fixed and another charged particle of mass m and charge q (of the same sign) is released from a distance r. The impulse of the force exerted by the external agent on the fixed charge by the time distance between Q and q becomes 2r is:
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Correct option is B)
As the other charge particle exerts force on the charge Q, the external agent has to apply equal and opposite force to keep it stationary. This force is thus equal to the force exerted on q by Q.
Net impulse is I=∫0tFdt=mΔvq
Initial Potential energy q is Ui=4πϵ01rQq
Final Potential energy of q is Uf=4πϵ012rQq
As the charge q is only under the influence of electrostatic forces, mechanical energy is conserved.