Question

A particle of mass M and charge q, initially at rest, is accelerated by a uniform electric field E through a distance D and is then allowed to approach a fixed static charge Q of the same sign. The distance of the closest approach of the charge q will then be:
A.$\dfrac{qQ}{4\pi {\epsilon}_{0}D}$
B.$\dfrac{Q}{4\pi {\epsilon}_{0}ED}$
C.$\dfrac{qQ}{2\pi {\epsilon}_{0}{D}^{2}}$
D.$\dfrac{Q}{4\pi {\epsilon}_{0}E}$

Answer 1

As the particle is initially at rest, its speed will be 0. Use the kinematics equation to find the final speed. Substitute this value in the equation of kinetic energy and calculate kinetic energy. Equate it with the potential energy of a charge and calculate r which is the distance of closest approach.

Complete answer:
Acceleration of particle is given by,
$a=\dfrac{F}{M}$…(1)
But, we know, $F=qE$
Substituting this value in eq. (1) we get,
$a=\dfrac{qE}{M}$
As the particle is initially at rest, its speed will be 0.
$\Rightarrow u=0$
Kinematics Equation is given by,
${v}^{2}-{u}^{2}=2aD$ …(2)
Substituting values in eq. (2) we get,
${v}^{2}-0=\dfrac{2qED}{M}$
$\Rightarrow {v}^{2}= \dfrac{2qED}{M}$
Now, we know Kinetic energy of a particle is given by,
$K.E.=\dfrac{1}{2}M{v}^{2}$
Substituting value of ${v}^{2}$ in above equation we get,
$K.E.=qED$
Potential energy of a particle is given by,
$P.E.= \dfrac{qQ}{4\pi {\epsilon}_{0}r}$
At closest approach, kinetic energy of a particle is equal to the potential energy.
$\Rightarrow K.E.=P.E.$
$\Rightarrow qED=\dfrac{qQ}{4\pi {\epsilon}_{0}r}$
Rearranging above equation we get,
$r=\dfrac{Q}{4\pi {\epsilon}_{0}ED}$
Thus, the distance of approach r=$\dfrac{Q}{4\pi {\epsilon}_{0}ED}$

Hence, the correct answer is option B i.e. $\dfrac{Q}{4\pi {\epsilon}_{0}ED}$.

Note:
When we say, distance of closest approach we refer to the distance between their centers. The nature of charge is not mentioned in the question. If the charge is positive then it will move in the same direction as the applied electric field. But, if the charge is negative, then it will be in the opposite direction to the applied electric field.