A particle of mass 'm' and charge 'q',initial | vedclass

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Question

(B) $1$. The work done by the uniform electric field '$E$' on the charge '$q$' over a distance '$D$' is equal to the kinetic energy gained by the part

Vedclass · Accepted Answer

$Q/4\pi\epsilon_0ED$

Vedclass · Answer

(B) $1$. The work done by the uniform electric field '$E$' on the charge '$q$' over a distance '$D$' is equal to the kinetic energy gained by the particle: $W = F \cdot D = qED$.$2$. Since the particle starts from rest, its kinetic energy at the end of distance '$D$' is $K = qED$.$3$. When the particle approaches the fixed charge '$Q$',it experiences a repulsive force. At the point of closest approach '$r$',the kinetic energy of the particle is completely converted into electrostatic potential energy.$4$. The electrostatic potential energy at distance '$r$' is given by $U = \frac{1}{4\pi\epsilon_0} \frac{qQ}{r}$.$5$. By the law of conservation of energy: $qED = \frac{1}{4\pi\epsilon_0} \frac{qQ}{r}$.$6$. Solving for '$r$': $r = \frac{Q}{4\pi\epsilon_0ED}$.

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