Advanced Integration: Variable Charge Densities

In our previous analysis of continuous bodies, we assumed the charge was spread uniformly, allowing us to treat the linear charge density $\lambda = Q/L$ as a constant. JEE Advanced routinely shatters this assumption.

When charge density is a function of position—such as $\lambda = f(x)$ or $\lambda = f(\theta)$—the setup of the differential element $dq$ must rigorously incorporate this function before any integration can occur.

Case 1: The Non-Uniform Rod

The Problem: A rod of length $L$ lies on the x-axis from $x = a$ to $x = a + L$. Its linear charge density varies proportionally with the distance from the origin, given by $\lambda = \lambda_0 x$, where $\lambda_0$ is a positive constant. Find the exact electrostatic force exerted on a positive point charge $q$ placed at the origin.

Step 1: Define the Differential Element

Consider an infinitesimal element of the rod of length $dx$ at a position $x$. The charge $dq$ on this element is no longer a simple fraction of the total charge. It is dictated by the local density:

$dq = \lambda \cdot dx = (\lambda_0 x) dx$

Step 2: Apply Coulomb's Law

Treating $dq$ as a point charge, the infinitesimal repulsive force $dF$ on the test charge $q$ (directed along the negative x-axis) is:

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

Substitute our expression for $dq$:

$dF = \frac{1}{4\pi\epsilon_0} \frac{q (\lambda_0 x \cdot dx)}{x^2} = \frac{q \lambda_0}{4\pi\epsilon_0} \frac{dx}{x}$

Notice the Mathematical Shift: Because the charge density grows linearly with $x$, it cancels one power of $x$ in the $1/x^2$ denominator of Coulomb's Law. This fundamentally changes the nature of the integral.

Step 3: Integration

We integrate $dF$ over the physical boundaries of the rod, from $x = a$ to $x = a + L$:

$F = \int_{a}^{a+L} \frac{q \lambda_0}{4\pi\epsilon_0} \frac{dx}{x}$

Pulling the constants out of the integral:

$F = \frac{q \lambda_0}{4\pi\epsilon_0} \int_{a}^{a+L} \frac{1}{x} dx$

The integral of $1/x$ is the natural logarithm, $\ln|x|$:

$F = \frac{q \lambda_0}{4\pi\epsilon_0} \left[ \ln(x) \right]_{a}^{a+L}$

$F = \frac{q \lambda_0}{4\pi\epsilon_0} \left( \ln(a+L) - \ln(a) \right)$

Applying logarithm properties ($\ln(A) - \ln(B) = \ln(A/B)$), we arrive at the final force:

$F = \frac{q \lambda_0}{4\pi\epsilon_0} \ln\left(1 + \frac{L}{a}\right)$

---

Case 2: The Semi-Circular Ring with Angular Variation

The Problem: A thin semi-circular ring of radius $R$ lies in the upper half of the xy-plane, centered at the origin. Its linear charge density varies with the angle $\theta$ (measured from the positive x-axis) according to the function $\lambda = \lambda_0 \cos\theta$. Find the force on a positive point charge $q$ located at the origin.

Step 1: The Geometry of the Differential Element

Let's take a small angular slice $d\theta$ at an angle $\theta$. The arc length of this slice is $ds = R d\theta$. The charge on this slice is:

$dq = \lambda \cdot ds = (\lambda_0 \cos\theta) (R d\theta)$

Step 2: The Force Vector and Symmetry

The force $dF$ from this element acts radially inward (since $\lambda$ is positive for $0 < \theta < \pi/2$ and $q$ is positive). The magnitude is:

$dF = \frac{1}{4\pi\epsilon_0} \frac{q \cdot dq}{R^2} = \frac{q \lambda_0 \cos\theta R d\theta}{4\pi\epsilon_0 R^2} = \frac{q \lambda_0}{4\pi\epsilon_0 R} \cos\theta d\theta$

We must resolve this into components:

• $dF_x = -dF \cos\theta$
• $dF_y = -dF \sin\theta$

The Symmetry Argument: The charge density is $\lambda_0 \cos\theta$. For $\theta$ between $0$ and $\pi/2$ (first quadrant), $\lambda$ is positive. For $\theta$ between $\pi/2$ and $\pi$ (second quadrant), $\cos\theta$ is negative, so $\lambda$ is negative. Because the first quadrant pushes the charge $q$ away, and the second quadrant pulls it in, the y-components will completely cancel out. We only need to integrate the x-component.

Step 3: Integrating the X-Component

$F_x = \int_{0}^{\pi} (-dF \cos\theta)$

Substitute our expression for $dF$:

$F_x = -\int_{0}^{\pi} \left( \frac{q \lambda_0}{4\pi\epsilon_0 R} \cos\theta \right) \cos\theta d\theta$

$F_x = -\frac{q \lambda_0}{4\pi\epsilon_0 R} \int_{0}^{\pi} \cos^2\theta d\theta$

Using the half-angle identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$:

$F_x = -\frac{q \lambda_0}{4\pi\epsilon_0 R} \int_{0}^{\pi} \left( \frac{1}{2} + \frac{\cos(2\theta)}{2} \right) d\theta$

$F_x = -\frac{q \lambda_0}{8\pi\epsilon_0 R} \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{0}^{\pi}$

Evaluating from $0$ to $\pi$, the sine terms vanish, leaving just $\pi$:

$F_x = -\frac{q \lambda_0 \pi}{8\pi\epsilon_0 R}$

The force is entirely along the negative x-axis. This problem demonstrates that understanding the physical symmetry dictated by the density function is just as critical as executing the integral.