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.

Gujarat GsebLong· 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.

Showing the 12 most recent of 16 on this concept.

GUJCET 2026Set x1 markMCQ
$q$
›Reveal solutionSolution
[!TLDR]

The numerator is the derivative of the denominator, so the integral is $\log|e^{x}+e^{-x}|+C$ — option (C).

Concept

Whenever an integrand has the form $\dfrac{f'(x)}{f(x)}$ , the integral is $\log|f(x)|+C$ . Here take $f(x)=e^{x}+e^{-x}$ , whose derivative is exactly $e^{x}-e^{-x}$ .

Solution

Let $u=e^{x}+e^{-x}$ . Then $du=(e^{x}-e^{-x})\,dx$ , and

$\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx=\int \frac{du}{u}=\log|u|+C=\log|e^{x}+e^{-x}|+C.$ …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. A point charge of 2.0 microC is at the centre of a cubic Gaussian surface 9.0 cm on edge. The net electric flux through the surface is ___ Nm^2/C.$
(a) 2.2 x 10^-6
(b) 2.2 x 10^5
(c) 2.2 x 10^6
(d) 2.2 x 10^-5
›Reveal solutionSolution

Gauss's law states the net electric flux through any closed surface is q_enclosed / epsilon_0, regardless of the surface's shape or size (as long as it encloses the same charge).

phi = q / epsilon_0

Given q = 2.0 microC = 2.0 x 10^-6 C, epsilon_0 = 8.85 x 10^-12 C^2/(N m^2).

phi = (2.0 x 10^-6) / (8.85 x 10^-12) = 2.26 x 10^5 N m^2/C

…

Start your 14-day free trial to see the full solution →
GUJCET 2025Set 031 markMCQ
$2q$
›Reveal solutionSolution
[!TLDR]

Using the shell theorem, $E' = 4kq/r^2 = 2E$ .

Concept

A uniformly charged thin spherical shell produces, at any external point, the same field as if all its charge were concentrated at the centre: $E = \dfrac{kQ}{d^2}$ .

Solution

For the point charge: $E = \dfrac{k(2q)}{r^2} = \dfrac{2kq}{r^2}$ . …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Consider a uniform electric field E = 3 x 10^3 î N/C. What is the flux of this field through a square of 10cm on a side whose plane is parallel to the xy plane?$
(a) 30 Nm^2/C
(b) Zero
(c) 15 Nm^2/C
(d) 60 Nm^2/C
›Reveal solutionSolution

Electric flux Φ = E·A = EA cosθ, where θ is the angle between the field and the surface's normal vector.

E = 3 × 10³ x̂ N/C is directed along x. The square lies in a plane parallel to the xy-plane, so its normal vector is along z — perpe …

Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. A charge q is placed at the center of one of the faces of a cube. The electric flux linked with the cube is ______.$
(a) q/ε0
(b) q/6ε0
(c) q/2ε0
(d) q/4ε0
›Reveal solutionSolution

Gauss's law gives total enclosed charge → flux, but here the charge sits exactly on a face, not fully inside the cube.

Imagine a second identical cube placed mirror-symmetric on the other side of that face, so together the two cubes fully enclose the charge q. By symmetry, each cube receives exactly half the total flux …

Start your 14-day free trial to see the full solution →
GUJCET 2024Set 131 markMCQ
$M^1L^3T^{-3}A^1$
›Reveal solutionSolution

Electric field has dimensions $MLT^{-3}A^{-1}$ ; multiplying by area $L^2$ gives electric flux $= M^1L^3T^{-3}A^{-1}$ .

Concept. Electric flux $\Phi_E = E\cdot A$ . Electric field $E=\dfrac{F}{q}$ has dimensions $\dfrac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$ . …
Start your 14-day free trial to see the full solution →
GUJCET 2024Set 131 markMCQ
$9 \times 10^4$
›Reveal solutionSolution

Using $E=\dfrac{2K\lambda}{r}$ , solve for $\lambda = \dfrac{Er}{2K}=10^{-7}$ C/m $=0.1\,\mu$ C/m.

Concept. The field of an infinite line charge is $E=\dfrac{2K\lambda}{r}=\dfrac{\lambda}{2\pi\varepsilon_0 r}$ .

Steps. With $E=9\times10^4$ N/C, $r=0.02$ m, $K=9\times10^9$ : …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. If an electric charge 'q' is placed at the centre of a cube, then the flux associated with each surface of the cube is ___.$
(a) q/ε0
(b) q/6ε0
(c) q/4ε0
(d) q/2ε0
›Reveal solutionSolution

By Gauss's law, total flux through a closed surface enclosing charge q is q/ε0; a cube has 6 identical faces symmetric about the centre.

Total flux through the cube (Gauss's law) = q/ε0.

…

Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. If two infinite plane sheets having same surface charge density σ are placed parallel to each other, then the electric field between the two sheets is ___.$
(a) zero
(b) σ/ε0
(c) σ/2ε0
(d) 2σ/ε0
›Reveal solutionSolution

Each infinite charged sheet produces a uniform field of magnitude σ/2ε0 pointing away from it (for positive σ) on both sides.

Between the two sheets, the field due to the left sheet points away from it (rightward, into the gap) with magnitude σ/2ε0, while the field due to the right sheet points away from it (leftward, into the gap) with the same magnitude σ/2ε0. Since both sheets carry the same sign and magnitude of charge d …

Start your 14-day free trial to see the full solution →
GUJCET 2023Set 091 markMCQ
$\vec{E} = 3 \times 10^3\,\hat{k}$
›Reveal solutionSolution
[!TLDR]

The field is along $\hat{k}$ while the area vector is along $\hat{i}$ , so the flux is zero.

Concept

Electric flux through a flat surface is $\Phi = \vec{E}\cdot\vec{A} = EA\cos\theta$ , where $\theta$ is the angle between the field and the outward normal (area vector).

Solution

The square lies in a plane parallel to the yz-plane, so its normal (area vector) is along the x-axis, $\vec{A} = A\,\hat{i}$ .

The field is $\vec{E} = 3\times10^{3}\,\hat{k}$ N/C. …
Start your 14-day free trial to see the full solution →
GUJCET 2023Set 091 markMCQ
$Q. Figure shows the electric field lines of four point charges A, B, C and D. [FIGURE: A has 3 field lines; B and C are joined by many field lines (dipole-like) with C also having outward lines; D has 4 field lines] Which charge has the maximum magnitude? (A) C charge (B) B charge (C) A charge (D) D charge$
›Reveal solutionSolution

Field-line count ∝ $|q|$ ; charge C has the most lines, so the largest magnitude.

Concept — field lines and charge magnitude. The number of electric field lines starting from (or ending on) a charge is proportional to the magnitude of that charge. Counting: A has 3 lines, D has 4 lines, while charges B and C are linked by many lines (a dipole-like pair) with C additionally showing outg …
Start your 14-day free trial to see the full solution →
GUJCET 2022Set 171 markMCQ
$M^1 L^{-3} T^{-3} A^{-1}$
›Reveal solutionSolution

$\Phi_E=E\cdot A$ ; with $[E]=MLT^{-3}A^{-1}$ and area $L^2$ , flux is $M^1L^3T^{-3}A^{-1}$ .

Concept: Electric field $E=\dfrac{\text{force}}{\text{charge}}=\dfrac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$ . …
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