Q.The electric field at a point is
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🔒 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 E passing through a small patch of area dA. The electric flux through that patch is:
dΦE=E⋅dA=EdAcosθ
where θ is the angle between E and the outward normal to the surface.
Why this definition?
- If E is perpendicular to the surface (θ=0), maximum field "flows through".
- If E is parallel (θ=90∘), no flux — the field just slides along the surface.
Total flux through a closed surface S is:
ΦE=∮SE⋅dA
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=4πε01r2q, radially outward.
- The area vector dA is also radially outward.
- So E⋅dA=EdA everywhere.
The total flux:
ΦE=∮EdA=E∮dA=(4πε01r2q)(4πr2)=ε0q
Notice: The r2 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 dΦ=4πε01r2qcosθdA.
- But r2cosθdA is exactly the solid angle dΩ subtended by dA at the charge.
So dΦ=4πε0qdΩ.
Integrating over the entire closed surface: ∮dΩ=4π (total solid angle around a point).
Hence:
ΦE=4πε0q⋅4π=ε0q
Result: For any closed surface enclosing q, the flux is ε0q.
4. Multiple Charges: Superposition
If there are many charges q1,q2,…,qn inside the surface, the total electric field is the vector sum of individual fields:
E=E1+E2+⋯+En
Flux is linear: …
The electrostatic field is a smooth function of position wherever there is no charge, but it changes abruptly where charge is present — across a charged sheet the normal field jumps by σ/ε0 and at a point charge it diverges. Options (a) and (c) are false, and while (b) is a true remark, the fact th …
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, ∮E⋅dA=qenc/ε0. 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 σ the normal component jumps by ΔE⊥=σ/ε0 (for example, just outside a charged conductor E=σ/ε0 while just inside E=0), and at a point charge the field diverges as 1/r2.
Checking the options.
- (a) is false: E is not continuous everywhere — it breaks at charges. …
Method: Superposition Principle for Electric Fields
Concept
The electric field at a point due to a system of charges is the vector sum of the fields produced by each charge individually. This is because electric fields obey the principle of linear superposition.
Steps
-
Identify all source charges
List every charge (q1,q2,…,qn) that contributes to the field at the point of interest.
-
Draw position vectors
From each charge to the point, draw the vector ri. The magnitude is the distance ri, and the direction is from the charge toward the point.
-
Write the field due to one charge
For a point charge qi, the electric field at distance ri is:
Ei=4πε01⋅ri2qir^i
where r^i is the unit vector pointing from the charge to the point.
- Repeat for every charge …
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 (E) with Electric Force (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 F=qE 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:
E=qF
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 E.
- Exam Tip: Always draw a diagram. Show the source charge and the point. Draw an arrow showing the direction of E.
Mistake 3: Using the Wrong Formula for the Source
- The Mistake: Using the formula for a point charge (E=r2kQ) 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=r2kQ (radially outward/inward)
- Infinite line charge: E=2πϵ0rλ (perpendicular to the line)
- Infinite plane sheet: E=2ϵ0σ (constant, perpendicular to the sheet)
- Conducting sphere (outside): E=r2kQ (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 θ with the x-axis:
Ex=Ecosθ
Ey=Esinθ
- **Add components separately:** $E_{net,x} = E_{1x} + E_{2x} + ...$ and $E_{net,y} = E_{1y} + E_{2y} + ...$ …
- KEAM 2026Set eng-2026-04194 marksMCQQ.If a spherical conductor of 10 cm radius contains 5×106 electrons, then the electric field on its surface (in NC−1) is (A) 0.86 (B) 0.36 (C) 0.45 (D) 1.44 (E) 0.72
›Reveal solutionSolution
Find the surface charge, then use E=kQ/r2 at the sphere's surface.
Total charge:
Q=ne=(5×106)(1.6×10−19)=8×10−13 C.
Field at the surface of a conducting sphere of radius r=0.1 m: …
- KEAM 2026Set eng-2026-04214 marksMCQQ.If an infinitely long uniformly charged wire produces an electric field of intensity E at a distance of d from it, then the linear charge density λ of the wire is (A) πϵ0Ed (B) 2πϵ0Ed (C) 21ϵ0Ed (D) 2πϵ0Ed (E) ϵ0Ed
›Reveal solutionSolution
Invert E=2πϵ0dλ to get λ=2πϵ0Ed.
For an infinite line charge, Gauss's law gives E=2πϵ0dλ at perpendicular distance d. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.A uniformly charged cube of side one cm having surface charge density of 8.85 μC cm−2 is placed inside a hollow metal sphere. The total flux emerging out of the sphere in Nm2C−1 is (ε0=8.85×10−12 C2N−1m−2) (A) 8.85×106 (B) 12×106 (C) 19.7×106 (D) 3×106 (E) 6×106
›Reveal solutionSolution
By Gauss's law the flux out of the enclosing sphere is the total charge divided by ε0.
The cube (side 1 cm) has 6 faces of 1cm2 each, total area 6cm2.
Total charge Q=σA=8.85μC cm−2×6cm2=53.1μC=53.1×10−6C. …
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.Electric-flux through a closed surface depends on the (A) shape of the surface (B) area of the surface (C) volume of the surface (D) electric field outside the surface (E) charge enclosed by the surface
›Reveal solutionSolution
Gauss's law: Φ=qenc/ε0 — only the enclosed charge matters.
Gauss's law states
ΦE=∮E⋅dA=ε0qenclosed. …
- KEAM 2025Set eng-2025-04234 marksMCQQ.The inward and outward electric flux from a closed surface are 6×104 NM2C−1 and 3×104 NM2C−1. Then the net charge (in coulomb) inside the closed surface is (A) −6×104ε0 (B) 6×104ε0 (C) 3×104ε0 (D) 9×104ε0 (E) [AMBIGUOUS]
›Reveal solutionSolution
[!TLDR]
Using Gauss's law with net flux =ϕout−ϕin=−3×104 gives an enclosed charge of −3×104ε0 C, which matches option (E).
Concept
Gauss's law states that the net electric flux through a closed surface equals the enclosed charge divided by the permittivity of free space: ϕnet=ε0qenc. Outward flux is taken positive and inward flux negative — the standard NCERT/CBSE electrostatics convention the KEAM syllabus is aligned with.
Solution
The net flux through the surface is the outward flux minus the inward flux:
ϕnet=ϕout−ϕin=3×104−6×104=−3×104 Nm2C−1.
By Gauss's law the enclosed charge is
qenc=ε0ϕnet=−3×104ε0 C. …
- KEAM 2025Set eng-2025-04274 marksMCQQ.The electric field inside a uniformly charged spherical shell of radius R is: (A) directly proportional to the charge within the shell (B) inversely proportional to R2 (C) same as that outside the shell (D) zero (E) maximum at the centre
›Reveal solutionSolution
A Gaussian surface inside a uniformly charged spherical shell encloses no charge, so E=0 everywhere inside.
For a uniformly charged spherical shell, apply Gauss's law to a concentric spherical surface of radius r<R. It encloses zero net charge, so …
- KEAM 2024Set eng-2024-06094 marksMCQQ.The electric field due to a an infinitely long thin wire with linear charge density λ at a radial distance r is proportional to (A) rλ2 (B) rλ (C) r2λ (D) rλ (E) rλ
›Reveal solutionSolution
By Gauss's law the field of an infinite line charge is E=2πε0rλ, i.e. proportional to λ/r.
Using a cylindrical Gaussian surface of radius r: …
- KEAM 2023Set eng-2023-P1-A14 marksMCQQ.A hollow sphere of radius 'r' encloses an electric dipole composed of two charges +q and −q. The net flux of electric field through the surface of the sphere due to the enclosed dipole is: (A) ε02q (B) ε02q⋅4πr2 (C) infinite (D) zero (E) ε0q
›Reveal solutionSolution
The net electric flux through the sphere is zero.
Concept and Intuition
By Gauss's law the flux depends only on the enclosed net charge. A dipole encloses +q and −q, whose sum is zero.
Step-by-Step Solution
- Enclosed charge =+q+(−q)=0.
- Gauss's law: Φ=ε0Qenc=ε00.
- Φ=0.
Common Mistakes …
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