Animated Solution for Physics - Electrostatics: Let $\rho (r)=\frac{Q}{\pi R^4}r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point $P$ inside the sphere at distance $r_1$ from the centre of the sphere, the magnitude of electric field is

When dealing with solid spheres in electrostatics, it is incredibly tempting to jump straight to the standard formula for the electric field inside a sphere: $E=\frac{kQr}{R^3}$. However, that formula comes with a massive caveat—it only works if the charge is distributed uniformly throughout the volume.

In this problem, we are thrown a curveball. The charge density is given by $\rho (r)=\frac{Q}{\pi R^4}r$. Notice that $r$ in the numerator? That means the charge density is zero at the exact center and grows linearly as you move outward toward the surface. To find the electric field at an internal point $P$ (at distance $r_1$), we must return to first principles and wield the full power of Gauss's Law.

Gauss's Law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space:

To exploit the spherical symmetry of the problem, we construct an imaginary spherical "Gaussian surface" of radius $r_1$ that passes exactly through point $P$. Because the charge distribution is spherically symmetric, the electric field $\vec{E}$ must point purely radially outward, and its magnitude must be constant everywhere on this Gaussian surface.

This symmetry beautifully simplifies the left side of Gauss's Law. The integral $\oint \vec{E}\cdot d\vec{A}$ simply becomes the magnitude of the electric field multiplied by the surface area of our Gaussian sphere:

The real heavy lifting in this problem is calculating $q_{enc}$, the total charge trapped inside our Gaussian surface of radius $r_1$. Because the density $\rho (r)$ is changing, we cannot just multiply density by volume. We must integrate.

Imagine the sphere is made up of infinitely many thin, concentric onion-like shells. Let's take one such shell at a generic radius $r$ with an infinitesimal thickness $dr$. The volume of this thin shell is its surface area multiplied by its thickness:

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

To find the total enclosed charge, we integrate these tiny charges from the center ($r=0$) all the way out to our Gaussian surface ($r=r_1$):

Let's clean up the constants. The $\pi$ in the numerator and denominator cancel out, and we can pull the constants outside the integral:

Now, we execute the simple power-rule integration:

We now have all the pieces of the puzzle. We substitute our flux and our enclosed charge back into Gauss's Law:

To isolate the electric field $E$, we divide both sides by $4\pi r_1^2$. Notice how the $r_1^2$ in the denominator cancels out two powers of $r_1$ in the numerator:

This is our final answer! It perfectly matches option (c).

A Pedagogical Takeaway: Notice how the electric field depends on the distance. For a uniform sphere, the internal field grows linearly ($E\propto r$). But here, because the charge density itself grows linearly with $r$, the resulting electric field grows quadratically ($E\propto r^2$). The physics perfectly reflects the mathematics.