Coulomb's Law for Continuous Charge Distributions

When charge is smeared over a physical object rather than concentrated at a geometric point, the standard scalar formulation of Coulomb's Law ($F = kq_1q_2/r^2$) cannot be applied directly to the macroscopic object. We must resort to the principles of calculus: slicing the object into infinitesimal point charges $dq$, and applying the principle of superposition via integration.

The Finite Charged Rod

Problem: A uniform thin rod of length $L$ carries a total positive charge $Q$. A point charge $q$ is placed on the axis of the rod at a distance $a$ from its near end. Calculate the exact electrostatic force on $q$.

Step 1: Define the Coordinate System and Charge Density

Let the point charge $q$ be at the origin $(0,0)$. The rod extends along the x-axis from $x = a$ to $x = a + L$. Because the charge is uniformly distributed along a line, we define the linear charge density $\lambda$:

$\lambda = \frac{Q}{L}$

Step 2: Isolate a Differential Element

Consider an infinitesimally small segment of the rod of length $dx$ located at an arbitrary position $x$. The charge contained in this segment is:

$dq = \lambda \cdot dx = \frac{Q}{L} dx$

Step 3: Apply Coulomb's Law to the Element

This differential element $dq$ acts as a point charge. The infinitesimal force $dF$ it exerts on $q$ is purely repulsive and directed along the negative x-axis:

$dF = \frac{1}{4\pi\epsilon_0} \frac{q \cdot dq}{x^2}$

Step 4: Integrate over the Limits of the Body

To find the total force $F$, we integrate $dF$ from the closest end of the rod to the furthest end:

$F = \int_{a}^{a+L} \frac{1}{4\pi\epsilon_0} \frac{q (\frac{Q}{L} dx)}{x^2}$

$F = \frac{qQ}{4\pi\epsilon_0 L} \int_{a}^{a+L} x^{-2} dx$

$F = \frac{qQ}{4\pi\epsilon_0 L} \left[ -\frac{1}{x} \right]_{a}^{a+L}$

$F = \frac{qQ}{4\pi\epsilon_0 L} \left( \frac{1}{a} - \frac{1}{a+L} \right)$

Simplifying the algebraic expression inside the parentheses yields the final rigorous result:

$F = \frac{1}{4\pi\epsilon_0} \frac{qQ}{a(a+L)}$

Sanity Check (The Point Charge Limit): What happens if the point charge is moved very far away, such that $a \gg L$? The term $a+L \approx a$, and the denominator becomes $a^2$. The formula collapses to $F = \frac{1}{4\pi\epsilon_0} \frac{qQ}{a^2}$, perfectly mirroring the force between two point charges. Checking limits is a crucial habit for complex JEE physics derivations.