The electric field at a point is (a) always continuous. (b) continuous if there is no charge at that point. (c) discontinuous only if there is a negative charge at that point. (d) discontinuous if there is a charge at that point.

The electric field at a point is (a) always continuous. (b) continuous if there is no charge at that point. (c

NCERT Exemplar · Q9

Q. The electric field at a point is

(a) always continuous.

(b) continuous if there is no charge at that point.

(d) discontinuous if there is a charge at that point.

CBSEMCQ· 1mImportance★★★★★

66% · 44/67 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 →

Field lines start and end on charges, so $E$ varies smoothly wherever there is no charge but changes abruptly (it jumps across a surface charge and diverges at a point charge) exactly where charge sits. Correct option: (d) discontinuous if there is a charge at that point.

Concept understanding. The electrostatic field obeys Gauss's law, $\oint \vec{E}\cdot d\vec{A}=q_{enc}/\varepsilon_0$ . In any charge-free region the field is a smooth, continuous function of position. A discontinuity appears only where charge is located: across a charged sheet of surface density $\sigma$ the normal component jumps by $\Delta E_\perp = \sigma/\varepsilon_0$ (for example, just outside a charged conductor $E=\sigma/\varepsilon_0$ while just inside $E=0$ ), and at a point charge the field diverges as $1/r^2$ .

Checking the options.

(a) is false: $E$ is not continuous everywhere — it breaks at charges. …

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

Here is a breakdown of the common mistakes students make when dealing with the concept of "The electric field at a point" , along with the correct understanding and exam-smart strategies to avoid them.

Mistake 1: Confusing Electric Field ( $\vec{E}$ ) with Electric Force ( $\vec{F}$ )

The Mistake: Students often think the electric field is the force. They might say "the field pushes the charge" without understanding the field is the cause of the force, not the force itself.
Why it happens: The formula $\vec{F} = q\vec{E}$ looks like they are the same thing, just multiplied by a charge.
How to Avoid:
- Remember the definition: The electric field at a point is the force per unit positive test charge placed at that point.
- Use the "test charge" logic: If you remove the test charge, the field still exists (created by other charges). The field is a property of space, not of the charge you place there.
- Exam Tip: If a question asks "What is the electric field at point P?" and gives you a force on a charge $q$ , do not say the field equals the force. Say:

\vec{E} = \frac{\vec{F}}{q}

Mistake 2: Forgetting the Vector Nature (Direction)

The Mistake: Students calculate the magnitude of the field correctly but forget to specify the direction or get the direction wrong (e.g., thinking the field points towards a positive charge).
Why it happens: Students treat electric field like a scalar (just a number) or confuse it with the direction of force on a negative charge.
How to Avoid:
- Memorize the rule: The electric field vector points away from a positive source charge and towards a negative source charge.
- Use the "positive test charge" rule: Imagine placing a tiny positive charge at the point. The direction of the force on that imaginary positive charge is the direction of $\vec{E}$ .
- Exam Tip: Always draw a diagram. Show the source charge and the point. Draw an arrow showing the direction of $\vec{E}$ .

Mistake 3: Using the Wrong Formula for the Source

The Mistake: Using the formula for a point charge ( $E = \frac{kQ}{r^2}$ ) when the source is a continuous charge distribution (like a line, ring, or sheet), or vice-versa.
Why it happens: Students memorize one formula and apply it everywhere without checking the geometry.
How to Avoid:
- Identify the source first. Is it a point charge? A dipole? An infinite line? A charged plate?
- Use the correct formula:
  - Point charge: $E = \frac{kQ}{r^2}$ (radially outward/inward)
  - Infinite line charge: $E = \frac{\lambda}{2\pi \epsilon_0 r}$ (perpendicular to the line)
  - Infinite plane sheet: $E = \frac{\sigma}{2\epsilon_0}$ (constant, perpendicular to the sheet)
  - Conducting sphere (outside): $E = \frac{kQ}{r^2}$ (same as point charge)
  - Conducting sphere (inside): $E = 0$
- Exam Tip: If the problem says "a charged rod" or "a charged ring," you cannot use the point charge formula directly. You must integrate or use a known result.

Mistake 4: Sign Errors in Superposition

The Mistake: When adding electric fields from multiple charges, students add magnitudes without considering that fields are vectors. They might treat a field pointing left as negative and a field pointing right as positive, but then add them incorrectly.
Why it happens: Students forget that electric field is a vector and superposition means vector addition.
How to Avoid:
- Always draw a coordinate system. Choose a clear x-axis and y-axis.
- Resolve each field into components. For example, if a field makes an angle $\theta$ with the x-axis:

E_x = E \cos\theta

E_y = E \sin\theta

- **Add components separately:** $E_{net,x} = E_{1x} + E_{2x} + ...$ and $E_{net,y} = E_{1y} + E_{2y} + ...$ …

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 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.

…

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

…

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

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

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

…

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

…

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

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

…

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

…

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

Start your 14-day free trial to see the full solution →
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 $^2$ …
Start your 14-day free trial to see the full solution →
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:

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

…

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

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

←Previous questionQ8 — If ointS vecE· dvecS = 0 over a surface, then (a) the electric field i…→Next questionQ10 — If there were only one type of charge in the universe, then (a) ointS …

Q.The electric field at a point is

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

Method: Superposition Principle for Electric Fields

Concept

Steps

Mistake 1: Confusing Electric Field ( $\vec{E}$ ) with Electric Force ( $\vec{F}$ )

Mistake 2: Forgetting the Vector Nature (Direction)

Mistake 3: Using the Wrong Formula for the Source

Mistake 4: Sign Errors in Superposition

Q.The electric field at a point is

🎓Unlock everything free for 14 days

Unlock everything free for 14 days