Animated Solution for Physics - Electrostatics: A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

Imagine you are given a solid sphere packed uniformly with charge, but someone has scooped out a smaller spherical chunk from inside it. What does the electric field look like inside that empty cavity? At first glance, it seems like a nightmare of integration. The boundaries are asymmetric, and the charge distribution is no longer simple. But physics offers us a beautiful, elegant shortcut: The Principle of Superposition.

Instead of dealing with the complex shape directly, we can imagine the system as a combination of two much simpler shapes. We pretend the original sphere is completely solid and intact, possessing a uniform positive charge density $+\rho$. Then, to account for the "missing" charge in the cavity, we superimpose a smaller solid sphere exactly at the cavity's location, but we give it a negative charge density $-\rho$.

When these two overlap, the $+\rho$ and $-\rho$ perfectly cancel each other out in the cavity region, leaving zero net charge—exactly matching our physical reality!

Let's define our geometry. Let the center of the large positive sphere be $O$, and the center of the small negative sphere (the cavity) be $O^{\prime }$. We define a vector $\vec{d}$ pointing from $O$ to $O^{\prime }$.

Now, pick any random point $P$ inside the cavity. We need to find the electric field here. Let the position vector of $P$ with respect to $O$ be $\vec{r}$, and its position vector with respect to $O^{\prime }$ be ${\vec{r}}^{\prime }$.

From the simple triangle law of vector addition, we can see that:

Which rearranges to:

We know that the electric field at a distance $\vec{r}$ inside a uniformly charged solid sphere is given by Gauss's Law as:

Using our superposition model, the net electric field ${\vec{E}}_{net}$ at point $P$ is simply the vector sum of the fields produced by the two individual spheres.

The field due to the large positive sphere is:

The field due to the small negative sphere is:

Notice the negative sign! It's crucial because this sphere carries a negative charge density.

Now, let's add them up:

Factoring out the common terms, we get:

And here is where the magic happens. Remember our vector triangle? We established that $\vec{r}-{\vec{r}}^{\prime }=\vec{d}$. Substituting this in, we arrive at our final, beautiful result:

Look closely at this equation. The vectors $\vec{r}$ and ${\vec{r}}^{\prime }$ have completely vanished! The net electric field depends only on $\vec{d}$, the vector connecting the centers of the two spheres. Since $\rho$, ${\epsilon }_0$, and $\vec{d}$ are all constants, the electric field inside the cavity is perfectly uniform and non-zero. It has the same magnitude and points in the exact same direction (parallel to $\vec{d}$) no matter where you are inside the cavity!