Answer

Verified

54k+ views

**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.

Recently Updated Pages

Why do we see inverted images in a spoon class 10 physics JEE_Main

A farsighted man who has lost his spectacles reads class 10 physics JEE_Main

State whether true or false The outermost layer of class 10 physics JEE_Main

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Other Pages

Which of the following distance time graph is representing class 11 physics JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

The angle between the hands of a clock when the ti-class-11-maths-JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main