Animated Solution for Physics - Electrostatics: Consider a sphere of radius $R$ which carries a uniform charge density $\rho$. If a sphere of radius $\frac{R}{2}$ is carved out of it, as shown in the figure the ratio $\frac{|{\mathbf{E}}_A|}{|{\mathbf{E}}_B|}$ of magnitude of electric field ${\mathbf{E}}_A$ and ${\mathbf{E}}_B$ respectively, at points $A$ and $B$ due to the remaining portion is

When dealing with cavities inside continuous charge distributions, the math can initially look terrifying. How do you integrate over a sphere with a chunk missing? The secret lies in the Principle of Superposition.

Instead of dealing with the complex geometry of a hollowed-out sphere, we can imagine the system as the sum of two perfect, complete spheres. First, we take a complete solid sphere of radius $R$ with a uniform positive charge density $\rho$. Then, we superimpose a smaller solid sphere of radius $R/2$ exactly where the cavity is, but we give it a negative charge density $-\rho$.

Where the two spheres overlap, the positive and negative charge densities perfectly cancel each other out ($\rho -\rho =0$), creating our empty cavity! This elegant trick allows us to calculate the electric field at any point by simply adding the field vectors from these two perfect spheres.

The Center of the Sphere

Let's start by finding the electric field at point $A$, which is the geometric center of the large solid sphere.

For the large positive sphere, point $A$ is exactly at its center. By symmetry, the electric field contributions from all infinitesimal charge elements around it cancel each other out perfectly. Therefore, the field due to the large sphere is zero:

Now, what about the smaller negative sphere (the cavity)? Point $A$ lies exactly on the surface of this smaller sphere. The formula for the electric field on the surface of a uniformly charged solid sphere is $E=\frac{\rho r}{3{\epsilon }_0}$. Substituting the cavity's radius $r=R/2$, we get:

Because the cavity has a negative charge density, this field points towards the center of the cavity (upwards). The net field at $A$ is simply the sum of these two contributions:

The Bottom Edge

Next, we move to point $B$, located at the bottom edge of the large sphere.

For the large positive sphere, point $B$ is on its surface. Using the same surface field formula, the field points radially outward (downwards) with a magnitude of:

Now, we must calculate the field at $B$ due to the negative cavity sphere. Point $B$ is completely outside the cavity. The distance from the center of the cavity to point $A$ is $R/2$, and the distance from $A$ to $B$ is $R$. Since they lie on the same vertical axis, the total distance $r$ from the cavity's center to $B$ is:

For a point outside a solid sphere, the electric field behaves as if all the charge were concentrated at the center: $E=\frac{\rho R_{cavity}^3}{3{\epsilon }_0{r}^2}$. Substituting our values:

Because the cavity is "negative," this field points towards the cavity (upwards). Since the two fields at $B$ point in opposite directions, we subtract their magnitudes to find the net field:

We have successfully found the magnitudes of the net electric field at both points. The final step is to find their ratio:

The $\rho R$ and ${\epsilon }_0$ terms cancel out beautifully, leaving us with pure fractions:

To match the given options, we multiply the numerator and denominator by 2:

This perfectly matches option (d). The principle of superposition has once again turned a complex geometric nightmare into a straightforward algebraic triumph!