Q.An early model for an atom considered it to have a positively charged point nucleus of charge $Ze$, surrounded by a uniform density of negative charge up to a radius $R$. The atom as a whole is neutral. For this model, what is the electric field at a distance $r$ from the nucleus?

The electric field can be determined using Gauss's Law, which is particularly effective for charge distributions with spherical symmetry.

1. Due to spherical symmetry, the electric field $\vec{E}$ is radial and its magnitude $E$ is constant on any spherical Gaussian surface centered at the nucleus. Thus, $\oint \vec{E} \cdot d\vec{A} = E(4\pi r^2)$.
2. The total negative charge in the atom is $-Ze$, uniformly distributed in a sphere of radius $R$. The volume charge density is $\rho = \frac{-Ze}{\frac{4}{3}\pi R^3} = \frac{-3Ze}{4\pi R^3}$.
3. We consider two cases for the enclosed charge $Q_{enc}$ within a Gaussian sphere of radius $r$:
   • Case 1: $r < R$ (inside the negative charge cloud). $Q_{enc} = Ze + \rho \left(\frac{4}{3}\pi r^3\right) = Ze + \left(\frac{-3Ze}{4\pi R^3}\right)\left(\frac{4}{3}\pi r^3\right) = Ze \left(1 - \frac{r^3}{R^3}\right)$.
   • **Case 2: $r \ge R$} (outside the atom). $Q_{enc} = Ze + (-Ze) = 0$, as the atom is neutral. …

The problem describes an early atomic model with a point nucleus of charge $Ze$ at the center, surrounded by a uniformly distributed negative charge cloud up to a radius $R$. The atom as a whole is neutral. We need to find the electric field at a distance $r$ from the nucleus.

This problem exhibits spherical symmetry. Whenever we have a spherically symmetric charge distribution, Gauss's Law is the most efficient tool to calculate the electric field. Gauss's Law states that the total electric flux through any closed surface (called a Gaussian surface) is proportional to the total electric charge enclosed within that surface.

Here, $\vec{E}$ is the electric field, $d\vec{A}$ is an infinitesimal area vector on the Gaussian surface, $Q_{enc}$ is the total charge enclosed by the Gaussian surface, and $\epsilon_0$ is the permittivity of free space. Due to spherical symmetry, the electric field $\vec{E}$ will be radial, pointing outwards for positive enclosed charge and inwards for negative enclosed charge. Its magnitude will be constant on any spherical surface centered at the nucleus. Thus, for a spherical Gaussian surface of radius $r$, the integral simplifies to $E(4\pi r^2)$.

We need to consider two distinct regions:

1. Outside the atom: $r > R$
2. Inside the negative charge cloud: $r < R$

Let's proceed step-by-step.

1. Determine the total negative charge and its volume charge density. The atom as a whole is neutral. Since the nucleus has a positive charge $Ze$, the total negative charge distributed in the cloud must be $-Ze$. This negative charge is uniformly distributed within a sphere of radius $R$. The volume of this sphere is $V = \frac{4}{3}\pi R^3$. The uniform volume charge density $\rho$ of the negative charge cloud is:

2. Calculate the electric field for $r > R$ (outside the atom). Consider a spherical Gaussian surface of radius $r$ such that $r > R$. This surface encloses the entire atom. The total charge enclosed, $Q_{enc}$, is the sum of the nucleus charge and the total negative charge cloud:

Applying Gauss's Law:

This result makes sense: a neutral atom, when viewed from a distance much larger than its size, appears as a point charge of zero magnitude, hence producing no external electric field.

3. Calculate the electric field for $r < R$ (inside the negative charge cloud).

Consider a spherical Gaussian surface of radius $r$ such that $r < R$.

The charge enclosed, $Q_{enc}$, by this Gaussian surface consists of two parts:

* The positive charge of the nucleus: $Ze$.

* The negative charge enclosed within the Gaussian sphere of radius $r$.

The volume of the Gaussian sphere is $V_r = \frac{4}{3}\pi r^3$.

The negative charge enclosed, $Q_{neg, enc}$, is the product of the charge density $\rho$ and this volume:

Now, the total charge enclosed by the Gaussian surface is: …

Method: Gauss’s Law with Spherical Symmetry

We use Gauss’s Law in integral form:

Because the charge distribution is spherically symmetric, the electric field is radial and depends only on $r$: $\vec{E} = E(r)\,\hat{r}$.

Step 1: Understand the charge distribution

• Positive point nucleus at centre: charge $+Ze$
• Negative charge cloud (uniform density) from $r=0$ to $r=R$, total negative charge = $-Ze$ (so atom is neutral)

Volume charge density of the negative cloud:

Step 2: Choose a Gaussian surface

A spherical surface of radius $r$, concentric with the nucleus.

Step 3: Find enclosed charge $Q_{\text{enc}}$ for two regions

Case 1: Inside the cloud ($r < R$)

Enclosed charge = nucleus charge + negative charge inside radius $r$:

Substitute $\rho$:

Simplify:

Case 2: Outside the cloud ($r \ge R$)

Atom is neutral, so total enclosed charge = $0$:

Step 4: Apply Gauss’s Law

Gauss’s Law gives: …

1. ✗ Forgetting that the atom is neutral overall

Mistake: Students often assume the total negative charge is $-Ze$ but then forget to check that the positive nucleus and negative cloud exactly cancel.

How to avoid:

Always state the neutrality condition explicitly:

Total positive charge $= +Ze$

Total negative charge $= -Ze$ (since atom is neutral)

This means the uniform negative charge density $\rho_{-}$ is:

Write this down before doing anything else.

2. ✗ Using the wrong Gaussian surface for $r < R$

Mistake: Students sometimes take a Gaussian sphere of radius $r$ but include all the negative charge (up to $R$) instead of only the part inside $r$.

How to avoid:

For $r < R$, the enclosed negative charge is only the charge inside radius $r$:

So the total enclosed charge is:

Key rule: Only charge inside the Gaussian surface counts.

3. ✗ Forgetting that the negative charge exists only up to $r = R$

Mistake: For $r > R$, students sometimes still include negative charge from the cloud — but outside $R$, the cloud doesn’t exist.

How to avoid:

For $r > R$:

• The entire negative charge ($-Ze$) is inside the Gaussian surface.
• So total enclosed charge $= +Ze - Ze = 0$.

Thus, for $r > R$, the electric field is zero.

4. ✗ Mixing up the direction of the field

Mistake: Students sometimes think the field points outward for all $r$, but inside the atom the field can point inward if the negative cloud dominates locally.

How to avoid:

Check the sign of $Q_{\text{enc}}$:

• For $r < R$: $Q_{\text{enc}} = Ze(1 - r^3/R^3)$ is positive (since $r < R$), so field points radially outward.
• For $r > R$: $Q_{\text{enc}} = 0$, so field is zero.

So the field is always outward (or zero), never inward.

5. ✗ Not applying Gauss’s law correctly for spherical symmetry

Mistake: Students write $\oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2$ but forget that this only works if $\vec{E}$ is radial and constant in magnitude on the Gaussian sphere.

How to avoid:

Always justify the symmetry: …