Question

A charged particle $q$ is shot with speed $v$ towards another fixed charged particle $Q$. It approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$, the closest distance of approach would be
A. $r$
B. $2r$
C. $\dfrac{r}{2}$
D. $\dfrac{r}{4}$

Answer 1

To find out the new closest distance of approach for the charged particle $q$ with speed $2v$, recall the concept of closest distance of approach. Use this concept to find the closest distance of approach for two different speeds of the charged particle $q$ and then compare them to get the required answer.

Complete step by step answer:
Given, the speed of the charged particle $q$ is $v$.Charge of a fixed particle is $Q$.Distance of closest approach is $r$.Let $r'$ be the closest distance of approach when the speed of the charged particle is $2v$.At closest distance of approach, the kinetic energy of the particle is converted to potential energy. That is here the kinetic energy of the charged particle $q$ will be converted into potential energy at closest distance of approach.
The formula for kinetic energy is written as,
$K.E = \dfrac{1}{2}m{v^2}$ (i)
where $m$ is the mass of the particle and $v$ is the velocity of the particle.
The potential energy between two charged particle is given the by the formula,
$U = \dfrac{{k{q_1}{q_2}}}{r}$ (ii)
where ${q_1}$ and ${q_2}$ are the charges of the two particles, $r$ is the distance between the two charges and $k$ is coulomb constant.

Now, when the speed of the charged particle $q$ is $v$ so kinetic energy will be (using equation (i)),
${\left( {K.E} \right)_1} = \dfrac{1}{2}m{v^2}$ (iii)
The potential energy at the closest distance of approach $r$ will be (using equation (ii)),
${U_1} = \dfrac{{kQq}}{r}$ (iv)
Now, equating the kinetic and potential energies at closest distance of approach $r$ we get,
${\left( {K.E} \right)_1} = {U_1}$
Putting the values of ${\left( {K.E} \right)_1}$ and ${U_1}$ we get,
$\dfrac{1}{2}m{v^2} = \dfrac{{kQq}}{r}$
$\Rightarrow r = \dfrac{{2kQq}}{{m{v^2}}}$ (v)

When the speed of the charged particle $q$ is $2v$, the kinetic energy will be (using equation (i)),
${\left( {K.E} \right)_2} = \dfrac{1}{2}m{\left( {2v} \right)^2}$ (vi)
The potential energy between the two charged particle at closest distance of approach $r'$ will be (using equation (ii))
${U_2} = \dfrac{{kQq}}{{r'}}$ (vii)
Now, equating the kinetic and potential energies at closest distance of approach $r'$ we get,
${\left( {K.E} \right)_2} = {U_2}$
Putting the values of ${\left( {K.E} \right)_2}$ and ${U_2}$ we get,
$\dfrac{1}{2}m{\left( {2v} \right)^2} = \dfrac{{kQq}}{{r'}}$
$\Rightarrow r' = \dfrac{{2kQq}}{{m4{v^2}}}$
$\Rightarrow r' = \dfrac{1}{4}\left( {\dfrac{{2kQq}}{{m{v^2}}}} \right)$ (viii)
Now substituting the value of $r$ from equation (v) in equation (viii) we get,
$\therefore r' = \dfrac{r}{4}$

Hence, the correct answer is option D.

Note: Remember the point that at distance of closest approach, the kinetic energy of the particle gets converted to potential energy. In problems involving closest distance of approach you will need to use this condition to find the required result. Also, remember the closest distance of approach is different for different speeds of the particle.