Consider a sphere of radius R with charge density – Class 12 Physics

NCERT Exemplar · Q27

R

(a) Find the electric field at all points

r

(b) Suppose the total charge on the sphere is

2e

where

e

is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.

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Concept understanding — Gauss Law

Gauss's Law is a shortcut. Instead of adding up the Coulomb contribution of every charge — a nightmare of vectors and integrals — it lets you read the field straight off the symmetry of a problem. The whole idea rests on one quantity: electric flux.

Flux — field-lines counted through a surface. For a uniform field E crossing a flat area A, the flux is Φ = E·A = EA cosθ, where θ is the angle between the field and the normal to the surface. Picture the field as a bundle of lines; flux is how many pierce the surface. When E is edge-on (θ = 90°) nothing passes through and Φ = 0; when E is ⊥ to the surface (θ = 0) the count is maximal. For a closed surface, Φ is the net number of lines leaving it — lines that enter and exit cancel.

The law itself. Gauss's law states that the net flux out of any closed surface depends only on the charge trapped inside:

Φ = q_enclosed / ε₀.

Two consequences do most of the work. First, only enclosed charge counts — a charge outside sends as many lines in as out, so its net contribution is exactly zero. Second, the surface's shape is irrelevant; move the charge around inside or deform the surface, and Φ never changes.

Why symmetry makes it powerful. By itself Φ = q/ε₀ has E buried in an integral. It becomes a tool only when you pick a Gaussian surface matched to the symmetry — one where E is constant and everywhere either ⊥ to the surface (so Φ = EA) or ∥ to it (contributing nothing). Then E slides out and you solve in one line. This works for exactly three geometries:

1 — Infinite line charge (linear density λ). Use a coaxial cylinder: E = λ / 2πε₀r, falling off as ∝ 1/r.

2 — Infinite sheet (surface density σ). Use a pillbox pierced through the sheet: E = σ / 2ε₀ — uniform and completely independent of distance. The field near a large charged plane simply doesn't weaken as you step back. …

Why this formula?

Gauss's Law: Why It Holds

Gauss's Law is one of the four Maxwell's equations and a cornerstone of electromagnetism. Let's build the understanding from the ground up — not just the formula, but the why.

1. The Core Idea: Flux as "Flow" of Field

Imagine an electric field $\vec{E}$ passing through a small patch of area $dA$ . The electric flux through that patch is:

d\Phi_E = \vec{E} \cdot d\vec{A} = E \, dA \cos\theta

where $\theta$ is the angle between $\vec{E}$ and the outward normal to the surface.

Why this definition?

If $\vec{E}$ is perpendicular to the surface ( $\theta = 0$ ), maximum field "flows through".
If $\vec{E}$ is parallel ( $\theta = 90^\circ$ ), no flux — the field just slides along the surface.

Total flux through a closed surface $S$ is:

\Phi_E = \oint_S \vec{E} \cdot d\vec{A}

2. The Key Insight: Flux Depends Only on Enclosed Charge

Consider a single point charge $+q$ at the centre of a spherical surface of radius $r$ .

By Coulomb's law, at every point on the sphere: $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$ , radially outward.
The area vector $d\vec{A}$ is also radially outward.
So $\vec{E} \cdot d\vec{A} = E \, dA$ everywhere.

The total flux:

\Phi_E = \oint E \, dA = E \oint dA = \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \right) (4\pi r^2) = \frac{q}{\varepsilon_0}

Notice: The $r^2$ cancels! The flux is independent of the radius.

3. Why Shape Doesn't Matter

Now, what if the surface is not spherical but any closed shape enclosing the charge?

Draw a small cone from the charge to the surface.
The flux through a patch $dA$ at distance $r$ is $d\Phi = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \cos\theta \, dA$ .
But $\frac{\cos\theta \, dA}{r^2}$ is exactly the solid angle $d\Omega$ subtended by $dA$ at the charge.

So $d\Phi = \frac{q}{4\pi\varepsilon_0} d\Omega$ .

Integrating over the entire closed surface: $\oint d\Omega = 4\pi$ (total solid angle around a point).

Hence:

\Phi_E = \frac{q}{4\pi\varepsilon_0} \cdot 4\pi = \frac{q}{\varepsilon_0}

Result: For any closed surface enclosing $q$ , the flux is $\frac{q}{\varepsilon_0}$ .

4. Multiple Charges: Superposition

If there are many charges $q_1, q_2, \dots, q_n$ inside the surface, the total electric field is the vector sum of individual fields:

\vec{E} = \vec{E}_1 + \vec{E}_2 + \dots + \vec{E}_n

Flux is linear: …

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Gauss's law gives $E=\dfrac{kr^2}{4\varepsilon_0}$ inside the sphere and $E=\dfrac{kR^4}{4\varepsilon_0 r^2}$ outside; the two protons must sit on opposite sides of the centre, each at $r_0=R/2^{3/4}=R/8^{1/4}$ , where the sphere's inward pull balances the outward proton–proton repulsion.

(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 $\rho=kr$ :

q(r)=\int_0^r (kr')\,(4\pi r'^2)\,dr' = 4\pi k\int_0^r r'^3\,dr' = 4\pi k\cdot\frac{r^4}{4}=\pi k r^4.

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

E=\frac{k r^2}{4\varepsilon_0}\qquad(r\le R),

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:

E=\frac{Q}{4\pi\varepsilon_0 r^2}=\frac{k R^4}{4\varepsilon_0 r^2}\qquad(r>R).

(b) Position of the two protons

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

q(r)=Q\frac{r^4}{R^4}=2e\,\frac{r^4}{R^4}.

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: …

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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:

\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}

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

E(r) \cdot 4\pi r^2 = \frac{Q_{\text{enc}}(r)}{\varepsilon_0}

(a) Finding the electric field everywhere

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

For $r \leq R$ :

Q_{\text{enc}}(r) = \int_0^r \rho(r') \cdot 4\pi r'^2 \, dr' = \int_0^r (k r') \cdot 4\pi r'^2 \, dr' = 4\pi k \int_0^r r'^3 \, dr'

Q_{\text{enc}}(r) = 4\pi k \cdot \frac{r^4}{4} = \pi k r^4

For $r > R$ :

Q_{\text{enc}} = \pi k R^4 \quad (\text{total charge of the sphere})

Step 2: Apply Gauss's Law

For $r \leq R$ :

E(r) \cdot 4\pi r^2 = \frac{\pi k r^4}{\varepsilon_0}

\boxed{E(r) = \frac{k r^2}{4\varepsilon_0}} \quad \text{(radially outward if k>0)}

For $r > R$ :

E(r) \cdot 4\pi r^2 = \frac{\pi k R^4}{\varepsilon_0}

\boxed{E(r) = \frac{k R^4}{4\varepsilon_0 r^2}} \quad \text{(same as point charge at origin)}

(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:

E(r) = \frac{k r^2}{4\varepsilon_0}

But $k$ is related to total charge:

Q = \pi k R^4 = 2e \quad \Rightarrow \quad k = \frac{2e}{\pi R^4}

So inside:

E(r) = \frac{2e \, r^2}{4\pi R^4 \varepsilon_0} = \frac{e \, r^2}{2\pi R^4 \varepsilon_0}

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). …

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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:

Q_{\text{enc}} = \rho \cdot \frac{4}{3}\pi r^3

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

✓ How to avoid:

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

Q_{\text{enc}} = \int_0^r \rho(r') \cdot 4\pi r'^2 \, dr'

For $\rho(r) = kr$ :

Q_{\text{enc}} = \int_0^r (k r') (4\pi r'^2) dr' = 4\pi k \int_0^r r'^3 dr' = 4\pi k \cdot \frac{r^4}{4} = \pi k r^4

🔍 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:

Q_{\text{enc}} = Q_{\text{total}} = \pi k R^4

Then:

E \cdot 4\pi r^2 = \frac{Q_{\text{total}}}{\varepsilon_0} \quad\Rightarrow\quad E = \frac{Q_{\text{total}}}{4\pi \varepsilon_0 r^2}

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:

Q_{\text{total}} = \pi k R^4 = 2e \quad\Rightarrow\quad k = \frac{2e}{\pi R^4}

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: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Higher Secondary (+2 Stage) Examination 2026Set ANNUAL1 markMCQ
$Q. An electric dipole formed by charges +q and -q separated by a distance 'd' is placed inside a hollow sphere of radius 'r' (2r > d). The electric flux through the surface of the sphere is —$
(a) q/ε0, outward
(b) 2q/ε0, inward
(c) 2q/ε0, outward
(d) zero
›Reveal solutionSolution

By Gauss's law, flux depends only on the net enclosed charge; a dipole has zero net charge, so the total flux through any closed surface enclosing it is zero.

By Gauss's theorem,

$\Phi_E = \frac{q_{enclosed}}{\varepsilon_0}$

Here the sphere of radius r (with 2r > d) encloses both charges of the dipole, +q and -q. The net enclosed charge is

$q_{enclosed} = (+q) + (-q) = 0$ …

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Higher Secondary (+2 Stage) Examination 2024Set ANNUAL1 markMCQ
$Q. A charge Q is enclosed by a spherical Gaussian surface of radius R. If the radius is doubled, the outward electric flux -$
(a) will decrease to half
(b) will remain the same
(c) will double
(d) will increase four-fold
›Reveal solutionSolution

Gauss's law says the outward electric flux through a closed surface depends only on the charge enclosed, never on the size or shape of the surface.

By Gauss's law:

$\Phi_E = \oint \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$

The enclosed charge is still $Q$ even after the radius is doubled to $2R$ — no charge has been added or removed, only the surface has grown. Since $\Phi_E$ depends only on $Q_{\text{enc}}$ and $\varepsilon_0$ (both unchanged), the flux is unaffected by the change in radius. (What …

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

(a) Electric field everywhere

(b) Position of the two protons

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

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