A metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$. A charge $Q$ is placed at the centre of the spherical cavity. What will be the surface charge density on (i) the inner surface, and (ii) the outer surface?

A metallic spherical shell has an inner radius R₁ and outer radius R₂. A charge Q is placed at the centre of t

NCERT Exemplar · Q15

R_1

(i) the inner surface, and

(ii) the outer surface?

Gujarat GsebShort· 2mImportance★★★★★

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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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Electrostatic induction forces charge $-Q$ onto the inner surface and $+Q$ onto the outer surface. The densities are $\sigma_{\text{inner}} = -\frac{Q}{4\pi R_1^2}$ and $\sigma_{\text{outer}} = +\frac{Q}{4\pi R_2^2}$ .

When a charge sits inside a conducting cavity, the conductor responds by rearranging its free electrons until equilibrium is reached. The key insight is that the electric field inside a conductor must be zero in electrostatic equilibrium. This constraint, combined with Gauss's law, tells us exactly how charge distributes on the surfaces.

The charge $Q$ at the center creates an electric field that would penetrate into the metal. But conductors won't allow that—electrons move until they cancel any internal field. This movement leaves one surface with a deficit of electrons (positive charge) and the other with an excess (negative charge).

Finding the charge distribution

1. Apply Gauss's law inside the conductor

Draw a Gaussian surface anywhere inside the metal itself (between $R_1$ and $R_2$ ). Since the electric field is zero throughout the conductor, the flux through this surface is zero:

\oint \vec{E} \cdot d\vec{A} = 0

By Gauss's law, this means the enclosed charge is zero:

Q_{\text{enclosed}} = \epsilon_0 \cdot 0 = 0

2. Determine the charge on the inner surface

The Gaussian surface encloses both the central charge $Q$ and whatever charge $q_{\text{inner}}$ sits on the inner surface at radius $R_1$ . For the total to be zero:

Q + q_{\text{inner}} = 0

Therefore:

q_{\text{inner}} = -Q

The inner surface must carry charge $-Q$ to neutralize the field inside the conductor.

3. Determine the charge on the outer surface

The spherical shell as a whole is electrically neutral (we started with an uncharged conductor). If the inner surface has $-Q$ , and the total charge on the shell is zero, then:

q_{\text{outer}} = +Q

The outer surface carries $+Q$ .

Tip

A quick check: the conductor had zero net charge initially, so $q_{\text{inner}} + q_{\text{outer}} = -Q + Q = 0$ . ✓

Computing the surface charge densities

4. Inner surface density

The charge $-Q$ spreads uniformly over the inner spherical surface of area $4\pi R_1^2$ : …

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Here are the common mistakes students make on this classic electrostatics problem, along with how to avoid each.

Mistake 1: Forgetting that charge resides only on the surfaces of a conductor

Many students assume the charge $Q$ at the centre somehow "spreads" throughout the shell material.

Why this is wrong: In electrostatic equilibrium, net charge inside a conductor is zero. The field inside the conducting material must be zero.

How to avoid:

Draw a Gaussian surface inside the conductor (between $R_1$ and $R_2$ ).
Since $E = 0$ inside the conductor, the total enclosed charge must be zero.
This forces the inner surface to acquire a charge of $-Q$ (to cancel the $+Q$ at the centre).

Mistake 2: Thinking the inner and outer surface charges are independent

Some students calculate the inner surface charge density correctly but then forget that the shell itself is neutral (unless stated otherwise).

Why this is wrong: The shell is uncharged overall. If the inner surface gets $-Q$ , the outer surface must get $+Q$ to keep the net charge zero.

How to avoid:

Always check conservation of charge for the conductor.
If the shell has a net charge $Q_{\text{net}}$ , then:

Q_{\text{outer}} = Q_{\text{net}} - Q_{\text{inner}}

In this problem, $Q_{\text{net}} = 0$ , so $Q_{\text{outer}} = +Q$ .

Mistake 3: Using the wrong area for surface charge density

Students often use the wrong radius when computing $\sigma = \frac{Q}{A}$ .

Common errors:

Using $R_2$ for the inner surface area.
Using $R_1$ for the outer surface area.
Forgetting that area is $4\pi r^2$ , not $\pi r^2$ .

How to avoid:

Inner surface area = $4\pi R_1^2$
Outer surface area = $4\pi R_2^2$
Then:

\sigma_{\text{inner}} = \frac{-Q}{4\pi R_1^2}

\sigma_{\text{outer}} = \frac{+Q}{4\pi R_2^2}

Mistake 4: Confusing the sign of the induced charge

Some students write $\sigma_{\text{inner}}$ as positive, thinking the central $+Q$ "repels" positive charge to the inner surface.

Why this is wrong: The central $+Q$ attracts negative charge to the inner surface (opposites attract). The induced charge on the inner surface is negative.

How to avoid:

Use the Gaussian surface argument: Inside conductor, $E = 0 \implies Q_{\text{enclosed}} = 0$ So $Q_{\text{inner}} + Q_{\text{centre}} = 0 \implies Q_{\text{inner}} = -Q$
The sign follows automatically from this condition.

--- …

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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 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

$\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx=\int \frac{du}{u}=\log|u|+C=\log|e^{x}+e^{-x}|+C.$ …
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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

…

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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: $E = \dfrac{k(2q)}{r^2} = \dfrac{2kq}{r^2}$ . …
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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 …

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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 …

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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 $\dfrac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$ . …
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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, $K=9\times10^9$ : …
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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.

…

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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 …

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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. …
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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 …
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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 $E=\dfrac{\text{force}}{\text{charge}}=\dfrac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$ . …
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Q.A metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$ . A charge $Q$ is placed at the centre of the spherical cavity. What will be the surface charge density on

Concept understanding — Gauss Law

Why this formula?

Gauss's Law: Why It Holds

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

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

3. Why Shape Doesn't Matter

4. Multiple Charges: Superposition

Finding the charge distribution

Computing the surface charge densities

Method: Electrostatic Induction in a Conductor (Gauss's Law + Charge Conservation)

Steps:

Mistake 1: Forgetting that charge resides only on the surfaces of a conductor

Mistake 2: Thinking the inner and outer surface charges are independent

Mistake 3: Using the wrong area for surface charge density

Mistake 4: Confusing the sign of the induced charge

Concept

Solution

Concept

Solution

Concept

Solution

Q.A metallic spherical shell has an inner radius R1R_1R1​ and outer radius R2R_2R2​. A charge QQQ is placed at the centre of the spherical cavity. What will be the surface charge density on

Finding the charge distribution

Computing the surface charge densities

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Q.A metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$ . A charge $Q$ is placed at the centre of the spherical cavity. What will be the surface charge density on

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