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.

Assam AhsecLong· 5mImportance★★★★★

93% · 62/67 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to unlock the full solution →

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). …

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to unlock the full solution →

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

AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL2 marks
$Q. Classify the following into$
(i) polar, and
(ii) non-polar molecules: CO2, HCl, H2 and H2O. OR Calculate the flux passing through a circular area of radius 5cm placed perpendicular to a uniform electric field E = 200i NC^-1.
›Reveal solutionSolution
Option 1: HCl and H₂O are polar (asymmetric charge distribution, net dipole moment); CO₂ and H₂ are non-polar (symmetric, zero net dipole moment). Option 2: electric flux $\Phi = EA \approx 1.57\,\mathrm{N\,m^2\,C^{-1}}$ .

Option 1 — Polar vs non-polar molecules

A molecule is polar if its centres of positive and negative charge do not coincide, giving it a permanent (net) dipole moment; it is non-polar if they coincide (net dipole moment zero), usually due to symmetry.

CO₂ — linear, symmetric ( $O=C=O$ ); the two C=O bond dipoles are equal and opposite, so they cancel $\Rightarrow$ non-polar.

HCl — different atoms (H, Cl) with unequal electronegativity, bond dipole does not cancel $\Rightarrow$ polar.

H₂ — homonuclear diatomic (identical atoms), no charge asymmetry $\Rightarrow$ non-polar.

H₂O — bent (angular) shape; the two O–H bond dipoles do not cancel $\Rightarrow$ polar.

So: Polar: HCl, H₂O. Non-polar: CO₂, H₂.

Option 2 — Electric flux

…
Start your 14-day free trial to see the full solution →
AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marks
$Q. A closed spherical surface encloses a charge q at its centre. Show that electric flux through the closed surface is q/ε0. OR A pair of charges +q and -q, separated by a small distance 2a is placed in an electric field E, so that the line joining the charges makes an angle θ with E. Write the expressions for torque τ and also its magnitude |τ|.$
›Reveal solutionSolution

Gauss's law shows the flux out of a sphere enclosing a point charge q is q/ε₀, independent of the sphere's radius; and the torque on a dipole in a field is τ = pE sinθ.

Main question: Consider a point charge $q$ at the centre O of a spherical surface of radius $r$ . By Coulomb's law, the magnitude of the electric field at every point on this sphere (all at distance $r$ from $q$ ) is the same:

$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

and it points radially outward, i.e. parallel to the outward area vector $d\vec A$ at every point on the sphere (by symmetry). So the flux through the whole closed surface is simply $E$ times the total surface area $4\pi r^2$ :

$\Phi_E = \oint \vec E \cdot d\vec A = E \times 4\pi r^2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\times 4\pi r^2 = \frac{q}{\epsilon_0}$

Notice the radius $r$ cancels out completely — the flux depends only on the enclosed charge $q$ , not on the size of the sphere. This is exactly Gauss's law for this special symmetric case.

…

Start your 14-day free trial to see the full solution →

←Previous questionQ26 — In 1959 Lyttleton and Bondi suggested that the expansion of the Univer…→Next questionQ28 — Two fixed, identical conducting plates α and β, each of surface area S…

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

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days