Question

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:

• A. $\sqrt{\frac{Qq}{4\pi {\in }_{0}mr}}$
• B. $\sqrt{\frac{Qqm}{4\pi {\in }_{0}r}}$
• C. $\sqrt{\frac{Qqm}{\pi {\in }_{0}r}}$
• D. $\sqrt{\frac{Qqm}{2\pi {\in }_{0}r}}$

Answer 1

Answer: (B)

$\sqrt{\frac{Qqm}{4\pi {\in }_{0}r}}$

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 $I={\int }_0^{t}Fdt=m\Delta {\overrightarrow{v}}_q$

Initial Potential energy $q$ is $U_i=\frac{1}{4\pi {ϵ}_0}\frac{Qq}{r}$

Final Potential energy of $q$ is $U_f=\frac{1}{4\pi {ϵ}_0}\frac{Qq}{2r}$

As the charge $q$ is only under the influence of electrostatic forces, mechanical energy is conserved.

Final Kinetic energy id $K_f=U_i-U_f=\frac{1}{4\pi {ϵ}_0}\frac{Qq}{2r}$

Thus, $v_q=\sqrt{\frac{2K_f}{m}}\Rightarrow I=m{v}_q=\sqrt{2m{K}_f}=\sqrt{\frac{Qqm}{4\pi {ϵ}_{0}r}}$