Animated Solution for Physics - Electrostatics: A charge $Q$ is uniformly distributed over a long rod $AB$ of length $L$ as shown in the figure. The electric potential at the point $O$ lying at distance $L$ from the end $A$ is

When dealing with point charges, calculating the electric potential is straightforward: we simply use the formula $V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$. However, when the charge is spread out over a continuous body—like a rod, a ring, or a sphere—we can no longer use this formula directly. The distance $r$ from our point of interest to the charge is not a single value; it varies depending on which part of the body we are looking at.

To solve this, we must invoke the power of calculus. We break the continuous body down into infinitesimally small elements. Each element is so small that it can be treated as a point charge. We then calculate the tiny potential $dV$ created by this tiny charge $dq$, and finally, we sum up (integrate) all these tiny potentials to find the total potential $V$.

In our specific problem, we have a rod $AB$ of length $L$ with a total charge $Q$ uniformly distributed over it. We need to find the electric potential at a point $O$, which lies on the same axis as the rod, at a distance $L$ from the nearer end $A$.

Let's set up a coordinate system. It's convenient to place the origin at point $O$. The rod then extends along the positive x-axis. The nearer end $A$ is at $x=L$, and the farther end $B$ is at $x=L+L=2L$.

We choose an infinitesimally small element of length $dx$ on the rod, located at a distance $x$ from the origin $O$.

Since the charge $Q$ is uniformly distributed over the length $L$, the linear charge density (charge per unit length) is:

The charge $dq$ contained within our small element $dx$ is simply the charge density multiplied by the length of the element:

Now, we can write the expression for the small potential $dV$ at point $O$ due to this element $dq$. Since the element is at a distance $x$ from $O$, we have:

Substituting our expression for $dq$ into this equation gives:

To find the total potential $V$, we must integrate $dV$ over the entire length of the rod. This means our integration variable $x$ will go from the position of end $A$ to the position of end $B$.

We can pull all the constant terms out of the integral:

The integral of $\frac{1}{x}$ with respect to $x$ is the natural logarithm, $\ln x$. Applying this, we get:

Now, we evaluate this at the upper and lower limits:

Using the logarithmic property $\ln a-\ln b=\ln \left(\frac{a}{b}\right)$, we can simplify the expression:

The $L$ terms inside the logarithm cancel out, leaving us with our final, elegant result:

This result beautifully demonstrates how calculus allows us to bridge the gap between discrete point charges and continuous physical objects.