Animated Solution for Physics - Electrostatics: Comprehension Passage

Welcome to this fascinating exploration of nuclear charge distribution! In this problem, we are given a nucleus of radius $R$ with a total charge of $Ze$. However, unlike a simple uniform sphere, the charge density $\rho (r)$ here has a specific profile: it remains constant at a value $d$ up to a distance $a$ from the center, and then it decreases linearly to zero at the surface $R$.

Let's break down the three questions based on this passage step by step.

Electric Field at the Surface

The first question asks for the electric field at the surface of the nucleus, where $r=R$.

To find this, we can elegantly apply Gauss's Law. Imagine a spherical Gaussian surface exactly at the boundary of the nucleus ($r=R$). According to Gauss's Law, the total electric flux through this surface is equal to the total enclosed charge divided by ${\epsilon }_0$.

Since our Gaussian surface encloses the entire nucleus, the total enclosed charge $q_{\text{in}}$ is simply the total nuclear charge, $Ze$. The electric field $E$ is uniform over the surface and points radially outward.

Notice something remarkable here? The expression for the electric field at the surface depends only on the total charge $Ze$ and the total radius $R$. It does not contain the parameter $a$ at all! This means that no matter how the charge is distributed internally (whether $a$ is small or large), as long as the total charge is $Ze$, the field at the surface remains the same. Therefore, the electric field at $r=R$ is independent of $a$.

Maximum Density when $a=0$

Now, let's consider a special case where $a=0$. If $a=0$, the region of constant density vanishes completely. The charge density $\rho (r)$ now starts at its maximum value $d$ at the center ($r=0$) and decreases linearly to zero at the surface ($r=R$).

The graph of $\rho (r)$ versus $r$ becomes a simple right-angled triangle. We can write the equation for this linear density profile as:

We need to find the value of this maximum density $d$. We know that the total charge of the nucleus is $Ze$. The total charge can also be found by integrating the charge density over the entire volume of the nucleus. Using spherical shells of volume $4\pi r^{2}dr$, we set up the integral:

Substitute our linear density function into the integral:

Let's pull out the constants and integrate term by term:

Plugging in the upper limit $R$:

Finally, solving for $d$, we get:

This gives us the maximum density at the center when the density profile is purely linear.

Linear Dependence of Electric Field

The final question presents an interesting observation: what if the electric field within the nucleus is linearly dependent on $r$? That is, $E\propto r$. What does this imply about our density profile?

Let's return to Gauss's Law for a point inside the nucleus at a distance $r$:

If we are given that $E$ is proportional to $r$ ($E=kr$ for some constant $k$), we can substitute this into Gauss's Law:

So, the enclosed charge must grow proportionally to the cube of the radius. But we also know that the enclosed charge is the volume integral of the density:

For the integral of $r^2$ to result in an $r^3$ term, the function $\rho (r)$ must be a constant. If $\rho (r)$ had any $r$ dependence (like our linear drop-off), the integral would produce higher powers of $r$ (like $r^4$).

Therefore, for the electric field to be strictly proportional to $r$ everywhere inside the nucleus, the charge density must be uniform throughout the entire volume.

Looking back at our original density profile, the density is constant only up to distance $a$. For it to be constant everywhere, the flat portion of the graph must extend all the way to the surface. This implies that $a=R$.

And there we have it! By carefully applying Gauss's Law and understanding volume integrals, we've successfully navigated through all the nuances of this nuclear charge distribution.