Animated Solution for Physics - Electrostatics: Let there be a spherically symmetric charge distribution with charge density varying as $\rho (r)={\rho }_0\left(\frac{5}{4}-\frac{r}{R}\right)$ upto $r=R$ and $\rho (r)=0$ for $r>R$, where $r$ is the distance from the origin. The electric field at a distance $r,(r<R)$ from the origin is given by

## Unraveling the Electric Field of a Non-Uniform Charge Distribution

Imagine a large sphere of radius $R$. Unlike a simple uniformly charged ball, the charge inside this sphere is distributed unevenly. The charge density $\rho (r)$ varies with the distance $r$ from the center according to the given function. Our mission is to find the electric field at a point inside this sphere, at a distance $r$ where $r<R$.

Visualizing the Sphere

To tackle this, we rely on the elegance of Gauss's Law. We construct an imaginary spherical Gaussian surface of radius $r$ concentric with our charge distribution.

Gauss's Law tells us that for a spherically symmetric charge distribution, the electric field $E$ at a distance $r$ depends only on the charge enclosed within that radius $r$. The charge outside this Gaussian surface exerts forces that perfectly cancel each other out inside.

Our master equation is:

Our entire problem now boils down to finding $q_{enc}$, the total charge trapped inside our Gaussian surface.

Finding the Enclosed Charge

Because the charge density $\rho (x)$ is not constant, we cannot simply multiply density by volume. Instead, we must use integration. Imagine peeling an onion; we divide the sphere into infinitesimally thin concentric shells.

Consider a thin shell at a distance $x$ from the center, with a tiny thickness $dx$. The volume of this shell is its surface area multiplied by its thickness:

The tiny amount of charge $dq$ contained in this shell is the density at that distance multiplied by the volume:

To find the total enclosed charge $q_{enc}$, we sum up all these tiny charges from the center ($x=0$) out to our Gaussian surface ($x=r$):

Adding up the Shells

Now, we substitute the given charge density function $\rho (x)={\rho }_0\left(\frac{5}{4}-\frac{x}{R}\right)$ into our integral:

We can pull the constants $4\pi$ and ${\rho }_0$ out of the integral to keep things clean:

Integrating term by term is straightforward. The integral of $x^2$ is $\frac{x^3}{3}$, and the integral of $x^3$ is $\frac{x^4}{4}$. Applying the limits from $0$ to $r$, we get:

Simplifying the fractions yields our enclosed charge:

Bringing it all Together

With $q_{enc}$ in hand, we return to our master equation from Gauss's Law:

The $4\pi$ terms beautifully cancel out.

To match the format of the given options, we factor out $r^3$ from the terms inside the parenthesis:

This simplifies to:

Finally, we factor out $\frac{1}{4}$ from the parenthesis to arrive at our ultimate expression:

This perfectly matches option (b). The beauty of this problem lies in the seamless transition from a physical concept (Gauss's Law) to a mathematical execution (volume integration), culminating in a clean, elegant result.