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.

Gujarat GsebMCQ· 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 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 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

Concept

Solution

Concept

Solution

Concept

Solution

Q.The electric field at a point is

🎓Unlock everything free for 14 days

Unlock everything free for 14 days