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# A positive point charge is released from rest at a distance ${r_0}$ from a positive line charge with uniform density. The speed $v$ of the point charge, as a function of instantaneous distance $r$ from line charge, is proportional toA. $v \propto {e^{ + r/{r_0}}}$B. $v \propto \ell n\left( {\dfrac{r}{{{r_0}}}} \right)$C. $v \propto \dfrac{r}{{{r_0}}}$D. $v \propto \sqrt {\ell n\left( {\dfrac{r}{{{r_0}}}} \right)}$

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Hint: You can apply conservation of energy in this system as there is no external force acting on the system.
When a point charge of mass $m$ is moving with some velocity $v$ , it possesses a kinetic energy which is given by $KE = \dfrac{1}{2}m{v^2}$ .
The electric field due to a line of charge at a distance $r$ is given by $E = \dfrac{\lambda }{{2\pi {\varepsilon _0}r}}$ where $\lambda$ is the linear charge density of the line of charge.
The potential energy at a point due to a point charge is the product of charge and electric potential at that point and relation between the potential and electric field is given by $dV = - \int {\vec E \cdot d\vec r}$ where $dV$ is the change in potential.

Complete step by step answer From the conditions given in the question it is clear that there is no external force acting on the system. So, we can apply conservation of energy on the system.
When a point charge of mass $m$ is moving with some velocity $v$ , it possesses a kinetic energy which is given by $KE = \dfrac{1}{2}m{v^2}$ .
Now, as the particle moves from ${r_0}$ to $r$ it must have done some work. And this work done will be the difference in the initial potential energy and final potential energy.
We know that the electric field due to a line of charge at a distance $r$ is given by $E = \dfrac{\lambda }{{2\pi {\varepsilon _0}r}}$ where $\lambda$ is the linear charge density of the line of charge.
The potential energy at a point due to a point charge is the product of charge and electric potential at that point and relation between the potential and electric field is given by $dV = - \int {\vec E \cdot d\vec r}$ where $dV$ is the change in potential.
So, the work done will be $W = - dV$
Now, the work done can be calculated after substituting the limits of integration as
$W = \int\limits_{{r_0}}^r {\dfrac{\lambda }{{2\pi {\varepsilon _0}r}}} dr = \dfrac{\lambda }{{2\pi {\varepsilon _0}}}\int\limits_{{r_0}}^r {\dfrac{{dr}}{r}}$
On simplifying the integration we have
$W = \dfrac{\lambda }{{2\pi {\varepsilon _0}}}\ell n\left( {\dfrac{r}{{{r_0}}}} \right)$
Now, we apply the conservation of energy and equate this work done with the kinetic energy of the point charge as
$\dfrac{1}{2}m{v^2} = \dfrac{\lambda }{{2\pi {\varepsilon _0}}}\ell n\left( {\dfrac{r}{{{r_0}}}} \right)$
So ignoring the constants we have
$v \propto \sqrt {\ell n\left( {\dfrac{r}{{{r_0}}}} \right)}$

Hence, option D is correct.

Note: The conservation of energy is basically a law which states that the energy is neither created nor destroyed although it can be converted into different forms. If a system is isolated from the surrounding i.e. there is no external force acting on the system, then the energy is conserved.
Last updated date: 31st May 2023
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