Animated Solution for Physics - Electrostatics: The electrostatic potential inside a charged spherical ball is given by $\varphi =a{r}^2+b$, where $r$ is the distance from the centre and $a,b$ are constants. Then, the charge density inside the ball is

Imagine you are exploring a mysterious charged spherical ball. You can't see the charges inside, but you have a map of the electrostatic potential $\varphi$ everywhere within it, given by the elegant equation $\varphi =a{r}^2+b$. Our mission is to reverse-engineer this potential to uncover the hidden volume charge density $\rho$ that creates it.

Finding the Electric Field

The first step in our journey is to find the electric field. The electric field is the spatial derivative—specifically, the negative gradient—of the potential. Since our potential only varies radially with distance $r$, the electric field $E$ is simply:

Substituting our potential function into the derivative:

Notice how the constant $b$ vanishes? It represents a uniform baseline potential that doesn't contribute to the electric field. The field grows linearly with distance from the center.

Uncovering the Enclosed Charge

Now that we have the electric field, we can invoke Gauss's Law to find the total charge $q$ enclosed within any radius $r$. Gauss's Law states that the electric flux through a closed surface equals the enclosed charge divided by ${\epsilon }_0$:

For a spherical Gaussian surface of radius $r$, the area is $4\pi r^2$. Plugging in our electric field:

Rearranging this gives us the enclosed charge as a function of $r$:

Slicing the Volume

We are almost there! We have the total charge $q$, but we want the volume charge density $\rho$, which is the charge per unit volume. Mathematically, $\rho =\frac{dq}{dV}$. Using the chain rule, we can split this into:

First, let's differentiate our charge $q$ with respect to $r$:

Next, we need the volume element $dV$ of a thin spherical shell of thickness $dr$, which is $dV=4\pi r^{2}dr$. Therefore, $\frac{dr}{dV}=\frac{1}{4\pi r^2}$. Multiplying these together yields our final answer:

The $r^2$ terms cancel out perfectly, revealing that the charge density inside the ball is completely uniform and constant!

The Elegance of Poisson's Equation

While the Gauss's Law method is highly intuitive, there is a faster, more advanced mathematical tool for this exact problem: Poisson's Equation, which directly relates potential to charge density:

In spherical coordinates, the Laplacian operator $ abla^2$ for a purely radial function is $\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d\varphi }{dr}\right)$. Let's apply it:

Equating this to $-\frac{\rho }{{\epsilon }_0}$:

In just three lines of calculus, we arrive at the exact same beautiful result!