A positive point charge is released from rest at a distance r0 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 to :-

v \propto \sqrt \ln ( r \over r_0 )

Electrostatics PYQ — JEE Main 2019 · 8 Apr · Shift 2 · Q65 with Solution

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Correct answer: D

Electric field due to an infinite line charge

For a line charge with uniform linear charge density $\lambda$ , the electric field at distance $r$ is

E(r)=\frac{\lambda}{2\pi \varepsilon_0 r}

Since the line charge is positive, a positive point charge placed near it is repelled outward.

Force on the point charge

If the point charge is $q$ , then the force at distance $r$ is

F(r)=qE(r)=\frac{q\lambda}{2\pi\varepsilon_0 r}

Use work-energy theorem

The charge is released from rest at $r_0$ , so initial kinetic energy is zero.

Work done by the electric force as it moves from $r_0$ to $r$ becomes its kinetic energy:

\frac{1}{2}mv^2 = \int_{r_0}^{r} F(r)\,dr

Substitute $F(r)$ :

\frac{1}{2}mv^2 = \int_{r_0}^{r} \frac{q\lambda}{2\pi\varepsilon_0 r}\,dr

\frac{1}{2}mv^2 = \frac{q\lambda}{2\pi\varepsilon_0} \int_{r_0}^{r} \frac{dr}{r}

\frac{1}{2}mv^2 = \frac{q\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r}{r_0}\right)

Solve for speed

v^2 \propto \ln\left(\frac{r}{r_0}\right)

Therefore,

v \propto \sqrt{\ln\left(\frac{r}{r_0}\right)}

Match with options

A: $v \propto \dfrac{r}{r_0}$ ❌
B: $v \propto \ln\left(\dfrac{r}{r_0}\right)$ ❌
C: $v \propto e^{r/r_0}$ ❌
D: $v \propto \sqrt{\ln\left(\dfrac{r}{r_0}\right)}$ ✅

So the correct option is D.

Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.

More from Electrostatics

View written solutionFree

Correct answer: D

Electric field due to an infinite line charge

For a line charge with uniform linear charge density $\lambda$ , the electric field at distance $r$ is

E(r)=\frac{\lambda}{2\pi \varepsilon_0 r}

Since the line charge is positive, a positive point charge placed near it is repelled outward.

Force on the point charge

If the point charge is $q$ , then the force at distance $r$ is

F(r)=qE(r)=\frac{q\lambda}{2\pi\varepsilon_0 r}

Use work-energy theorem

The charge is released from rest at $r_0$ , so initial kinetic energy is zero.

Work done by the electric force as it moves from $r_0$ to $r$ becomes its kinetic energy:

\frac{1}{2}mv^2 = \int_{r_0}^{r} F(r)\,dr

Substitute $F(r)$ :

\frac{1}{2}mv^2 = \int_{r_0}^{r} \frac{q\lambda}{2\pi\varepsilon_0 r}\,dr

\frac{1}{2}mv^2 = \frac{q\lambda}{2\pi\varepsilon_0} \int_{r_0}^{r} \frac{dr}{r}

\frac{1}{2}mv^2 = \frac{q\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r}{r_0}\right)

Solve for speed

v^2 \propto \ln\left(\frac{r}{r_0}\right)

Therefore,

v \propto \sqrt{\ln\left(\frac{r}{r_0}\right)}

Match with options

A: $v \propto \dfrac{r}{r_0}$ ❌
B: $v \propto \ln\left(\dfrac{r}{r_0}\right)$ ❌
C: $v \propto e^{r/r_0}$ ❌
D: $v \propto \sqrt{\ln\left(\dfrac{r}{r_0}\right)}$ ✅

So the correct option is D.

Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.

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