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.

Showing the 12 most recent of 37 on this concept.

CBSE 2026Set A1 markMCQ
$Q. S.I. unit of electric flux is (A) Vm (B) Vm^2 (C) Jm (D) NC^-1$
›Reveal solutionSolution

Φ = E·A → (V/m)(m²) = V·m.

Electric flux is $\Phi = \vec{E}\cdot\vec{A}$ .

Unit of electric field $E$ = N/C = V/m (volt per metre).

Unit of area $A$ = m².

…

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CBSE 2026Set A1 markMCQ
$Q. The surface charge densities on the surface of two conducting spheres of radii r1 and r2 are equal. The ratio of electric field intensities on the surfaces is (A) r1/r2 (B) r1^2/r2^2 (C) r2^2/r1^2 (D) 1 : 1$
›Reveal solutionSolution

Just outside a charged conductor E = σ/ε₀; with equal σ the fields are equal (1:1).

The electric field just outside the surface of a charged conductor is

$E = \frac{\sigma}{\varepsilon_0},$

which depends only on the local surface charge density σ, not on the radius.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Electric flux is a$
(a) scalar quantity
(b) vector quantity
(c) scalar or vector quantity
(d) constant quantity
›Reveal solutionSolution

Electric flux is a scalar quantity, even though it is defined using two vectors.

Electric flux through a surface is defined as

$\Phi = \oint \vec{E}\cdot d\vec{A}$

Although $\vec{E}$ (electric field) and $d\vec{A}$ (area vector, normal to the surface element) are both vectors, their dot product $\vec{E}\cdot d\vec{A} = E\,dA\cos\theta$ is a single number (magnitude only, with a sign depending on $\theta$ ) — it …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A charge Q, is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will$
(a) decrease to half
(b) increase two times
(c) remain unchanged
(d) increase four times
›Reveal solutionSolution

Gauss's law: flux through any closed surface = Q_enclosed / epsilon_0, and this does NOT depend on the surface's size or shape.

Gauss's law states that for any closed (Gaussian) surface,

flux (phi) = Q_enclosed / epsilon_0

Here the same charge Q sits at the centre of the sphere both before and after the radius is doubled - the enclosed charge Q_enclosed is unchanged. Since flux depends ONLY on Q_enclosed and the permittivity of free space epsilon_0 (both unchanged here), the flux does not change even though the surface area (4piR^2) has increased fourfold. Doubling R spreads the same total flux over 4 ti …

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CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) increases as $r$ increases for $r<R$ and for $r>R$
(b) is zero as $r$ increases for $r<R$ and decreases as $r$ increases for $r>R$
(c) is zero as $r$ increases for $r<R$ and increases as $r$ increases for $r>R$
(d) decreases as $r$ increases for $r<R$ and for $r>R$
›Reveal solutionSolution

A charged conducting (hollow metal) sphere carries all its charge on the outer surface. Gauss's law gives $E=0$ inside and $E\propto 1/r^2$ (decreasing) outside.

Setting up Gauss's law

For a hollow, uniformly charged conducting sphere of radius $R$ and total charge $Q$ , all the charge resides on the outer surface (a fundamental property of conductors in electrostatic equilibrium — free charges repel each other and move to the surface where the electric field inside the conducting material is zero).

Take a concentric spherical Gaussian surface of radius $r$ .

Case 1: $r<R$ (inside the shell)

The Gaussian surface of radius $r$ encloses no charge, because all the charge $Q$ lies on the surface at radius $R>r$ .

$\oint \vec E\cdot d\vec A = \frac{Q_{\text{enc}}}{\varepsilon_0} = 0 \implies E=0$

This is true for every $r<R$ — the field is zero throughout the interior, it does not "increase" or "decrease," it is simply zero.

Case 2: $r>R$ (outside the shell)

Now the Gaussian surface encloses the entire charge $Q$ . By spherical symmetry, $\vec E$ is radial and has the same magnitude everywhere on the Gaussian sphere, so

…

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CBSE 2026Set ANNUAL1 mark
$Q. Match the Column A with Column B and write the correct pair. Column A: Electric flux. Column B:$
(i) μ₀nI,
(ii) Volt × second,
(iii) μ₀nI/2,
(iv) Volt × meter,
(v) Volt × meter⁻¹,
(vi) Volt.
›Reveal solutionSolution

Electric flux φ_E = E·A; unit = (V/m)(m²) = V·m, option (iv).

Electric flux through a surface is φ_E = E·A (for a uniform field perpendicular to area A). The SI unit of electric field E is volt per metre (V/m) = N/C, and area is in m². Therefore the unit of electric flux is

…

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CBSE 2025Set D1 markMCQ
$Q. Gauss's law states that the electric flux through a closed surface is (A) proportional to the charge enclosed (B) inversely proportional to the charge enclosed (C) zero (D) proportional to the square of the charge enclosed$
›Reveal solutionSolution

Gauss's law states the total electric flux through a closed surface equals the enclosed charge divided by ε₀, so Φ ∝ q_enclosed.

Gauss's law is written as

$\oint \vec{E}\cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}$

The left side is the total electric flux Φ through the closed (Gaussian) surface. Thus

$\Phi = \frac{q_{enc}}{\varepsilon_0}$

…

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CBSE 2025Set D1 markMCQ
$Q. Inside a closed surface n electric dipoles are situated. The electric flux coming out from the closed surface will be (A) q/ε0 (B) 2q/ε0 (C) nq/ε0 (D) zero$
›Reveal solutionSolution

A dipole has zero net charge, so n dipoles enclose zero charge and the net flux is zero.

Gauss's law states the net electric flux out of a closed surface is Φ = q_enclosed/ε₀.

Each electric dipole consists of +q and −q; its net charge is +q + (−q) = 0. With n dipoles inside, the total enclosed charge is n × 0 = 0.

…

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CBSE 2025Set A1 mark
$Q. Write True or False: Inside a conductor, electrostatic field is zero.$
›Reveal solutionSolution

The statement is True: the electrostatic field inside a conductor is zero in equilibrium.

In electrostatic equilibrium, free charges in a conductor redistribute themselves on the surface such that the electric field inside the body of the conductor is exactly zero. If there were a residual field inside, it would exert a force on the free electrons, causing them to keep moving — contradicting the assumption …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The SI unit of electric flux is$
(a) N C m^2
(b) N C^-1 m^-2
(c) N C^-1 m^2
(d) N^2 C^-1 m^2
›Reveal solutionSolution
Electric flux is the field strength times the perpendicular area through which it passes, so its unit is simply the product of the units of $E$ and area.

Electric flux through a surface is defined as $\Phi_E = \vec{E}\cdot\vec{A}$ (or $\int \vec{E}\cdot d\vec{A}$ for a general surface).

SI unit of electric field $E$ = N C $^{-1}$

SI unit of area $A$ = m $^2$ …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The SI unit of electric flux is:$
(a) $Nm^2C^{-2}$
(b) $NC^{-1}m^2$
(c) $CN^2m^{-1}$
(d) $C^2N^{-1}m^{-2}$
›Reveal solutionSolution

Electric flux $\Phi_E = \vec{E}\cdot\vec{A}$ , so its unit is the unit of $E$ (N C⁻¹) times the unit of area (m²).

Electric flux through a surface is defined as $\Phi_E = \oint \vec{E}\cdot d\vec{A}$ , i.e. the product of the electric field and the area component perpendicular to it.

Since the SI unit of electric field $E$ is newton per coulomb ( $NC^{-1}$ ) and the unit of area $A$ is square metre ( $m^2$ ), the unit of electric flux is:

$NC^{-1} \times m^2 = NC^{-1}m^2$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The SI unit of the surface integral of electric field is -$
(a) Vm
(b) V
(c) $NC^{-1}$
(d) $Cm^{-3}$
›Reveal solutionSolution

Electric flux $\Phi_E=\oint\vec{E}\cdot d\vec{A}$ has SI unit V·m (equivalent to N·m $^2$ /C).

The surface integral of the electric field, $\Phi_E = \oint \vec{E}\cdot d\vec{A}$ , is the electric flux. Since $E$ has SI unit V/m (or equivalently N/C) and area has unit m $^2$ , the flux has uni …

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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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