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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
$\sqrt {\frac{{Qq}}{{4\pi { \in _0}mr}}} $
$\sqrt {\frac{{Qqm}}{{4\pi { \in _0}r}}} $
$\sqrt {\frac{{Qqm}}{{\pi { \in _0}r}}} $
$\sqrt {\frac{{Qqm}}{{2\pi { \in _0}r}}} $
Solution
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 $I=\int_{0}^{t} F d t=m \Delta \vec{v}_{q}$
Initial Potential energy $q$ is $U_{i}=\frac{1}{4 \pi \epsilon_{0}} \frac{Q q}{r}$
Final Potential energy of $q$ is $U_{f}=\frac{1}{4 \pi \epsilon_{0}} \frac{Q q}{2 r}$
As the charge $q$ is only under the influence of electrostatic forces, mechanical energy is conserved.
Final Kinetic energy id $K_{f}=U_{i}-U_{f}=\frac{1}{4 \pi \epsilon_{0}} \frac{Q q}{2 r}$
Thus, $v_{q}=\sqrt{\frac{2 K_{f}}{m}} \Rightarrow I=m v_{q}=\sqrt{2 m K_{f}}=\sqrt{\frac{Q q m}{4 \pi \epsilon_{0} r}}$
