Q.Consider a sphere of radius $R$ with charge density distributed as $\rho(r) = kr$ for $r \leq R$ and $\rho(r) = 0$ for $r > R$.

1. Field from Gauss's law. For a Gaussian sphere of radius $r \le R$, the enclosed charge is

   $q(r)=\int_0^r (kr')\,4\pi r'^2\,dr' = 4\pi k\int_0^r r'^3\,dr' = \pi k r^4.$

   So $E\,(4\pi r^2)=\dfrac{\pi k r^4}{\varepsilon_0}$, giving

   $E=\dfrac{k r^2}{4\varepsilon_0}\quad(r\le R),\ \text{radially outward.}$

   For $r>R$ the whole charge $Q=\pi k R^4$ is enclosed, so

   $E=\dfrac{kR^4}{4\varepsilon_0 r^2}\quad(r>R).$

2. Where the protons go. With total charge magnitude $2e$, the charge inside radius $r$ is $q(r)=2e\,(r/R)^4$. Place the two protons symmetrically on opposite sides of the centre, each at radius $r_0$ (separation $2r_0$). The sphere's field pulls each proton inward while the other proton pushes it outward; equating magnitudes, …

(a) Electric field everywhere

The distribution is spherically symmetric, so $\vec E$ is radial and depends only on $r$. Choose a concentric spherical Gaussian surface.

Inside ($r\le R$). The charge enclosed is the volume integral of $\rho=kr$:

Gauss's law $E\,(4\pi r^2)=q(r)/\varepsilon_0$ gives

directed radially outward for $k>0$. The field grows as $r^2$ (not linearly) because the density itself increases with $r$.

Outside ($r>R$). The full charge $Q=\pi k R^4$ is enclosed, and the sphere acts like a point charge:

(b) Position of the two protons

Take the total charge magnitude as $Q=2e$, so the charge within radius $r$ is

By symmetry the two protons must lie on a diameter, one on each side of the centre at the same radius $r_0$, a distance $2r_0$ apart. Each proton feels two radial forces: …

Method: Gauss's Law with Spherical Symmetry

Concept: For spherically symmetric charge distributions, the electric field is radial and depends only on the enclosed charge. Gauss's Law states:

For a spherical Gaussian surface of radius $r$, this simplifies to:

(a) Finding the electric field everywhere

Step 1: Find total charge $Q_{\text{enc}}(r)$ inside radius $r$

For $r \leq R$:

For $r > R$:

Step 2: Apply Gauss's Law

For $r \leq R$:

For $r > R$:

(b) Position for zero force on embedded protons

Step 1: Relate $k$ to total charge

Given total charge $Q = 2e$ (positive, since protons are positive and the sphere is negative).

The problem states total charge is $2e$ and protons are embedded. For force on each proton to be zero, the net electric field at that point must be zero.

Since the sphere has negative charge distribution and total charge is $2e$ (positive), the sphere must have net positive charge. The protons experience repulsion from the sphere's positive charge.

Step 2: Condition for zero force

For a proton at radius $r$, the electric field inside the sphere is:

But $k$ is related to total charge:

So inside:

Step 3: Where can field be zero?

Inside the sphere, $E(r) > 0$ for $r > 0$. Outside, $E(r) > 0$ for all $r$. The only point where $E = 0$ is at $r = 0$ (the centre). …

Great — this is a classic JEE/NEET problem on non-uniform charge distributions and Gauss’s law. Let’s go through the common mistakes systematically.

🔍 Common Mistake #1: Forgetting that $\rho$ is not constant

Students often treat $\rho = kr$ as if it were uniform and write:

This is wrong because $\rho$ depends on $r$.

✓ How to avoid:

Always use the integral form for enclosed charge when $\rho$ is not constant:

For $\rho(r) = kr$:

🔍 Common Mistake #2: Using the wrong Gaussian surface for $r > R$

Some students apply Gauss’s law with a sphere of radius $r > R$ but forget that outside the sphere, the charge enclosed is the total charge, not $\pi k r^4$.

✓ How to avoid:

For $r > R$, the enclosed charge is fixed:

Then:

This is exactly the field of a point charge $Q_{\text{total}}$ at the centre — a key sanity check.

🔍 Common Mistake #3: Forgetting to find $k$ from total charge

Part (b) gives total charge $Q_{\text{total}} = 2e$. Students sometimes try to solve without linking $k$ to $2e$.

✓ How to avoid:

Always compute $k$ explicitly:

Then use this $k$ in the expression for $E(r)$ inside the sphere.

🔍 Common Mistake #4: Misinterpreting “force on each proton is zero”

Students think this means the protons must be at the centre (where $E=0$). But inside a non-uniform sphere, $E=0$ only at $r=0$.

✓ How to avoid:

For a proton to feel zero net force, the electric field at its position must be zero. Since both protons are positive, they repel each other — so they cannot both be at $r=0$.

The only way both have zero force is if:

• They are placed symmetrically about the centre.
• The net field from the sphere plus the other proton cancels at each location.

This leads to solving: …