Two large charged plane sheets of charge densities $\sigma$ and $-2\sigma$ C/m$^2$ are arranged vertically with a separation of $d$ between them. Deduce expressions for the electric field at points (i) to the left of the first sheet, (ii) to the right of the second sheet, and (iii) between the two sheets. OR A spherical conducting shell of inner radius $r_1$ and outer radius $r_2$ has a charge $Q$. (a) A charge $q$ is placed at the centre of the shell. Find out the surface charge density on the inner and outer surfaces of the shell. (b) Is the electric field inside a cavity (with no charge) zero; independent of the fact whether the shell is spherical or not ? Explain.

> [!TLDR] > Sheet fields (rightward $+$): (i) $+\sigma/2\varepsilon_0$, > (ii) $-\sigma/2\varepsilon_0$, > (iii) $+3\sigma/2\varepsilon_0$. Shell with central $q$: inner density $-q/4\pi r_1^2$, outer $(Q+q)/4\pi r_2^2$; an empty cavity has zero field for any shape. ## Part (a) — superposition of two infinite sheets An infinite sheet of density $\sigma$ produces a uniform field $E=\dfrac{|\sigma|}{2\varepsilon_0}$ on each side — directed **away** from a positive sheet and **toward** a negative sheet. Sheet 1 ($+\sigma$) gives $\dfrac{\sigma}{2\varepsilon_0}$; sheet 2 ($-2\sigma$) gives $\dfrac{2\sigma}{2\varepsilon_0}=\dfrac{\sigma}{\varepsilon_0}$. Take rightward as $+$, with sheet 1 on the left and sheet 2 on the right. - **(i) Left of sheet 1.** Sheet 1's field points left ($-\sigma/2\varepsilon_0$); sheet 2 (negative) pulls toward itself, i.e. right ($+\sigma/\varepsilon_0$). Net: $$E_i=-\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{\varepsilon_0}=+\frac{\sigma}{2\varepsilon_0}\ \text{(rightward)}.$$ - **(ii) Right of sheet 2.** Sheet 1's field points right ($+\sigma/2\varepsilon_0$); sheet 2 pulls toward itself, i.e. left ($-\sigma/\varepsilon_0$). Net: $$E_{ii}=+\frac{\sigma}{2\varepsilon_0}-\frac{\sigma}{\varepsilon_0}=-\frac{\sigma}{2\varepsilon_0}\ \text{(leftward)}.$$ - **(iii) Between the sheets.** Sheet 1 points right ($+\sigma/2\varepsilon_0$); sheet 2 pulls toward itself, again right ($+\sigma/\varepsilon_0$). Net: $$E_{iii}=+\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{\varepsilon_0}=+\frac{3\sigma}{2\varepsilon_0}\ \text{(rightward)}.$$ …

Two large charged plane sheets of charge densities σ and -2σ C/m² are arranged vertically with a separation of

Question

\sigma

(i) to the left of the first sheet,

(ii) to the right of the second sheet, and

(iii) between the two sheets.

(OR)

r_1

(a) A charge

q

is placed at the centre of the shell. Find out the surface charge density on the inner and outer surfaces of the shell.

(b) Is the electric field inside a cavity (with no charge) zero; independent of the fact whether the shell is spherical or not ? Explain.

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Part (a)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: …

Part (b)Concept understanding — Electrostatic Induction

Electrostatic Induction: From Intuition to Precision

Imagine you're holding a charged balloon near a wall. The balloon sticks to the wall, even though the wall is uncharged. How does that happen? That's electrostatic induction at work.

The Core Intuition

Every material is made of atoms, which contain equal numbers of positive protons and negative electrons. Normally, these charges are balanced — the material is neutral. But charges can rearrange inside a material without leaving it.

When you bring a charged object near a neutral conductor, the charges inside that conductor feel a force. Like charges repel, opposite charges attract. So the free electrons in the conductor shift slightly: they move toward a positive external charge, or away from a negative one. The conductor as a whole stays neutral, but one side becomes slightly positive and the other slightly negative. This separation of charge is electrostatic induction.

Note

Induction does not involve contact. The charged object never touches the conductor — it only comes near. The effect is purely due to the electric field.

The Precise Statement

Electrostatic induction is the redistribution of electric charge in an object, caused by the influence of a nearby charged body, without direct contact between them.

Key points to remember:

The inducing charge remains unchanged.
The induced charges are equal in magnitude but opposite in sign on the two sides of the conductor.
The total charge on the induced object remains zero (if it was initially neutral).
The effect is strongest in conductors (where charges move freely), but also occurs in insulators (where molecules polarise slightly).

A Simple Demonstration

Take a metal sphere (uncharged) and bring a positively charged rod near it — not touching.

Electrons in the metal are attracted toward the rod. The side nearest the rod becomes negatively charged.
The far side, now lacking electrons, becomes positively charged.
The sphere is still neutral overall, but charge has been induced.

If you now touch the far side of the sphere with your finger while the rod is still near, electrons flow from your body into the sphere (or out of it, depending on sign). Remove your finger first, then remove the rod — the sphere now has a net charge. This is charging by induction, a direct application. …

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Sheet fields (rightward $+$ ): (i) $+\sigma/2\varepsilon_0$ ,

(ii) $-\sigma/2\varepsilon_0$ ,

(iii) $+3\sigma/2\varepsilon_0$ . Shell with central $q$ : inner density $-q/4\pi r_1^2$ , outer $(Q+q)/4\pi r_2^2$ ; an empty cavity has zero field for any shape.

Part (a) — superposition of two infinite sheets

An infinite sheet of density $\sigma$ produces a uniform field $E=\dfrac{|\sigma|}{2\varepsilon_0}$ on each side — directed away from a positive sheet and toward a negative sheet. Sheet 1 ( $+\sigma$ ) gives $\dfrac{\sigma}{2\varepsilon_0}$ ; sheet 2 ( $-2\sigma$ ) gives $\dfrac{2\sigma}{2\varepsilon_0}=\dfrac{\sigma}{\varepsilon_0}$ . Take rightward as $+$ , with sheet 1 on the left and sheet 2 on the right.

(i) Left of sheet 1. Sheet 1's field points left ( $-\sigma/2\varepsilon_0$ ); sheet 2 (negative) pulls toward itself, i.e. right ( $+\sigma/\varepsilon_0$ ). Net:

E_i=-\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{\varepsilon_0}=+\frac{\sigma}{2\varepsilon_0}\ \text{(rightward)}.

(ii) Right of sheet 2. Sheet 1's field points right ( $+\sigma/2\varepsilon_0$ ); sheet 2 pulls toward itself, i.e. left ( $-\sigma/\varepsilon_0$ ). Net:

E_{ii}=+\frac{\sigma}{2\varepsilon_0}-\frac{\sigma}{\varepsilon_0}=-\frac{\sigma}{2\varepsilon_0}\ \text{(leftward)}.

(iii) Between the sheets. Sheet 1 points right ( $+\sigma/2\varepsilon_0$ ); sheet 2 pulls toward itself, again right ( $+\sigma/\varepsilon_0$ ). Net: $E_{iii}=+\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{\varepsilon_0}=+\frac{3\sigma}{2\varepsilon_0}\ \text{(rightward)}.$ …

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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 38 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.Two large charged plane sheets of charge densities $\sigma$ and $-2\sigma$ C/m $^2$ are arranged vertically with a separation of $d$ between them. Deduce expressions for the electric field at points

Part (a)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

Part (b)Concept understanding — Electrostatic Induction

Electrostatic Induction: From Intuition to Precision

The Core Intuition

The Precise Statement

A Simple Demonstration

Field of charged sheets; conductor with a cavity

Part (a) — two charged plane sheets

Part (a) — superposition of two infinite sheets

Q.Two large charged plane sheets of charge densities σ\sigmaσ and −2σ-2\sigma−2σ C/m2^22 are arranged vertically with a separation of ddd between them. Deduce expressions for the electric field at points

Part (a) — superposition of two infinite sheets

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Q.Two large charged plane sheets of charge densities $\sigma$ and $-2\sigma$ C/m $^2$ are arranged vertically with a separation of $d$ between them. Deduce expressions for the electric field at points

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