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 $2 r$ is:

Question

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Answer 1

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 $I = \int_{0}^{t} F d t = m Δ v_{q}$

Initial Potential energy $q$ is $U_{i} = \frac{1}{4 π ϵ _{0}} \frac{Q q}{r}$
Final Potential energy of $q$ is $U_{f} = \frac{1}{4 π ϵ _{0}} \frac{Q q}{2 r}$

As the charge $q$ is only under the influence of electrostatic forces, mechanical energy is conserved.
Final Kinetic energy id $K_{f} = U_{i} - U_{f} = \frac{1}{4 π ϵ _{0}} \frac{Q q}{2 r}$

Thus, $v_{q} = \frac{2 K _{f}}{m} \Rightarrow I = m v_{q} = 2 m K_{f} = \frac{Q q m}{4 π ϵ _{0} r}$

Answer 2

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 Δ v_{q}

Initial Potential energy

q

is

U_{i} = \frac{1}{4 π ϵ _{0}} \frac{Q q}{r}

Final Potential energy of

q

is

U_{f} = \frac{1}{4 π ϵ _{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 π ϵ _{0}} \frac{Q q}{2 r}

Thus,

v_{q} = \frac{2 K _{f}}{m} \Rightarrow I = m v_{q} = 2 m K_{f} = \frac{Q q m}{4 π ϵ _{0} r}