A positive point charge is released from rest at a distance r0 from a positive line charge with unif

Q: Why can't we just use the equation $v^2 = u^2 + 2as$ to find the final velocity?

That's a very common trap! The equations of kinematics are strictly valid only when the acceleration is constant. Here, the electric force (and thus acceleration) depends on the distance $x$, meaning it decreases as the charge moves away. We must use integration to account for this variable force.

Q: Why is the electric field inversely proportional to $x$ and not $x^2$ like a point charge?

Great question! A point charge spreads its field lines in three dimensions (a sphere), leading to an inverse-square law. A line charge, however, spreads its field lines in two dimensions (a cylinder). By Gauss's Law, the surface area of a cylinder grows linearly with radius $x$, leading to an $1/x$ dependence.

Q: What would happen if the line charge was negative instead of positive?

If the line charge were negative, the electric field would point inwards. Our positive charge $q$ would be attracted towards the wire instead of being repelled. The setup would be similar, but the limits of integration and the sign of work done would reflect an inward motion.

Q: Why do we use the natural logarithm ($\ln$) instead of base-10 logarithm ($\log$)?

In calculus, the integral of $1/x$ with respect to $x$ naturally yields the natural logarithm, denoted as $\ln(x)$ or $\log_e(x)$. Base-10 logarithm is primarily used for scaling in engineering (like decibels or pH), but nature's math always defaults to base $e$.

Q: Does the mass of the charge affect this proportionality?

No, it doesn't! While a heavier mass would certainly result in a lower absolute speed $v$ for the same amount of work done, the mass $m$ is just a constant in our final equation. When we look for proportionality with respect to the distance $r$, all constants (including mass, charge, and $\lambda$) are absorbed, leaving only the mathematical function of $r$.

Question

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 to

Options:

(A) $v \propto \left(\frac{r}{r_0}\right)$

(B) $v \propto e^{+\frac{r}{r_0}}$

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

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

Sigma Percentile · Accepted Answer

### The Setup: A Repulsive Push

Imagine you are holding a tiny positive charge $q$ near an infinitely long, positively charged wire. The wire has a uniform linear charge density $\lambda$. Because like charges repel, the moment you let go of the charge from its initial distance $r_0$, it will accelerate outwards.

Our mission is to find out how its speed $v$ depends on its distance $r$ from the wire at any given instant. To do this, we must first understand the environment the charge is moving through. By applying Gauss's Law to a cylindrical surface around the wire, we know that the electric field $E$ at any radial distance $x$ is given by:

$$E = \frac{\lambda}{2\pi\epsilon_0 x}$$

### The Variable Force and Work Done

As the charge moves, it experiences an electrostatic force $F = qE$. Substituting our electric field expression, the force becomes:

$$F = \frac{q\lambda}{2\pi\epsilon_0 x}$$

Notice a crucial detail here: the force is **not constant**. It is inversely proportional to the distance $x$. Because the force changes continuously as the particle moves, we cannot rely on our standard constant-acceleration kinematic equations (like $v^2 = u^2 + 2as$). Instead, we must use the power of calculus to find the work done.

For an infinitesimally small displacement $dx$, the tiny amount of work done $dW$ by the electric field is simply $F \, dx$. To find the total work done $W$ as the charge moves from its starting point $r_0$ to a new distance $r$, we integrate this expression:

$$W = \int_{r_0}^{r} \frac{q\lambda}{2\pi\epsilon_0 x} dx$$

Pulling the constants out of the integral, we are left with the integral of $1/x$, which evaluates to the natural logarithm $\ln(x)$:

$$W = \frac{q\lambda}{2\pi\epsilon_0} [\ln x]_{r_0}^{r} = \frac{q\lambda}{2\pi\epsilon_0} (\ln r - \ln r_0)$$

Using the properties of logarithms, we can condense this beautifully into:

$$W = \frac{q\lambda}{2\pi\epsilon_0} \ln\left(\frac{r}{r_0}\right)$$

### The Work-Energy Connection

Now, we ask ourselves: what does this work actually *do*? According to the **Work-Energy Theorem**, the net work done on an object translates directly into a change in its kinetic energy ($\Delta K$).

Since the charge was released from rest, its initial kinetic energy was zero. Therefore, the final kinetic energy is simply $\frac{1}{2}mv^2$. Equating the work done to this kinetic energy gives us our master equation:

$$\frac{1}{2}mv^2 = \frac{q\lambda}{2\pi\epsilon_0} \ln\left(\frac{r}{r_0}\right)$$

We are looking for a proportionality relationship for the speed $v$. If we strip away all the constants (like mass $m$, charge $q$, $\lambda$, and $\epsilon_0$), we can clearly see how $v^2$ depends on the spatial variables:

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

Taking the square root of both sides reveals the final, elegant truth of the particle's motion:

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

This tells us that while the particle continues to speed up as it moves away, its rate of acceleration drops drastically, governed by the slow growth of the square root of a natural logarithm.

Animated Solution for Physics - Electrostatics: 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 to

Select Answer:

Visualized Solution

Visualizing the Setup

Electric Field of a Line Charge

Force on the Charge

Work Done for Small Displacement

Setting up the Integral

Evaluating the Integral

Work-Energy Theorem

Final Proportionality

The Sigma Insight: Electric Field

The Setup

The Variable Force and Work Done

The Work-Energy Connection

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