Question

A charged particle is shot from a large distance towards another charged particle which is fixed, with a speed. It approaches up to a closest distance and then returns. If were given a speed, the distance of approach would be

          

(a)$r$
(b)$2r$
(c)$\dfrac{r}{2}$
(d)$\dfrac{r}{4}$

Answer 1

Hint:In order to calculate If $q$ were given a speed \[2v\], the distance of approach, formula used as mentioned below. We will solve it by taking different cases.

Complete answer
For first case,
change in potential energy at closest distance is\[\dfrac{kQq}{r}\]
kinetic energy is $\dfrac{m{{v}^{2}}}{2}$
By conservation of energy,
\[\dfrac{kQq}{r}=\dfrac{m{{v}^{2}}}{2}\]
$\Rightarrow r=\dfrac{2kQq}{m{{v}^{2}}}$……….(i)
In second case,
when charge is launched with velocity \[2v\], let closest distance be ${{r}_{1}}$
similar to above case,
By conservation of energy,
\[\dfrac{kQq}{{{r}_{1}}}=\dfrac{m{{\left( 2v \right)}^{2}}}{2}\]
$\Rightarrow {{r}_{1}}=\dfrac{2kQq}{m{{v}^{2}}}$
From equation (1),
${{r}_{1}}=\dfrac{r}{4}$

Hence, the answer to this question is option (d).

Additional Information: Potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors. Electric potential energy, or Electrostatic potential energy, is a potential energy (measured in joules) that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system.

Note:While solving this question, we should be aware of the different types of formula used here. The formulae are basically from potential energy of a system of charges and how the different values of the variable of the formula is used from the question. The formula is modified and used here to take out the required solution for the problem given here. Different formulae are used here which must be taken into consideration while solving the question.