A charged particle of charge Q is held fixed | vedclass

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Question

(B) To keep the charge $Q$ fixed,an external agent must apply a force equal and opposite to the electrostatic force exerted by $q$ on $Q$. By Newton's

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$\sqrt{\frac{Qqm}{4\pi \epsilon_0 r}}$

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(B) To keep the charge $Q$ fixed,an external agent must apply a force equal and opposite to the electrostatic force exerted by $q$ on $Q$. By Newton's third law,this force is equal in magnitude to the force exerted by $Q$ on $q$. The impulse $I$ on the external agent is equal to the change in momentum of the particle $q$.By the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy of particle $q$:$U_i - U_f = K_f - K_i$Since $K_i = 0$:$K_f = \frac{1}{4\pi \epsilon_0} \frac{Qq}{r} - \frac{1}{4\pi \epsilon_0} \frac{Qq}{2r} = \frac{Qq}{8\pi \epsilon_0 r}$Using $K_f = \frac{1}{2}mv_q^2$,we find the velocity $v_q$:$v_q = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2}{m} \cdot \frac{Qq}{8\pi \epsilon_0 r}} = \sqrt{\frac{Qq}{4\pi \epsilon_0 mr}}$The impulse $I$ is the change in momentum of particle $q$:$I = m \cdot v_q = m \sqrt{\frac{Qq}{4\pi \epsilon_0 mr}} = \sqrt{\frac{Qqm}{4\pi \epsilon_0 r}}$

$A$ 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 the distance between $Q$ and $q$ becomes $2r$ is:

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