Q.A cube has side length a. A point charge q is placed, in four separate cases, at the following positions relative to the cube:
(a) at point A, which is a corner (vertex) of the cube;
(b) at point B, the mid-point of one edge of the cube;
(c) at point C, the centre of one face of the cube;
(d) at point D, the mid-point of the straight segment joining B and C (so D lies on that face of the cube). In each case find the total electric flux through all the faces of the cube.
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 angledΩ 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:
Surround the charge with enough identical cubes to fully enclose it, then share the total flux q/ε0 equally. A corner is shared by 8 cubes, an edge-midpoint by 4, a face-point by 2. So the fluxes are q/8ε0, q/4ε0, q/2ε0 and q/2ε0.
By Gauss's law the flux from a fully-enclosed charge is q/ε0. When the charge sits on a symmetry point shared by several identical cubes, that total flux divides equally among them: a corner is shared by 8 cubes, an edge-midpoint by 4, and any point lying on a face by 2. This directly gives the four answers.
Concept
Gauss's law: a charge fully enclosed gives total flux Φtot=q/ε0. If the charge lies on a boundary shared by n identical cubes tiling the space around it, symmetry splits the flux equally, so each cube gets Φ=nε0q.
Method: The Symmetry-Sharing Trick for Flux Through a Partial Enclosure
Use this whenever a point charge sits exactly ON the boundary of a closed surface — a corner, edge, or face — so the surface alone does not fully enclose it.
Steps
Step 1: Recall that a charge fully enclosed by ANY closed surface gives total flux q/ε0 (Gauss's law), independent of the surface's shape.
Step 2: Mentally tile identical copies of the given surface around the charge's location until the charge IS fully enclosed by the group.
The number of copies needed equals how many such surfaces meet symmetrically at that exact point: 8 cubes meet at a shared corner, 4 meet along a shared edge, 2 meet across a shared face. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL2 marks
Q.Classify the following into
(i) polar, and
(ii) non-polar molecules: CO2, HCl, H2 and H2O.
OR
Calculate the flux passing through a circular area of radius 5cm placed perpendicular to a uniform electric field E = 200i NC^-1.
›Reveal solutionSolution
Option 1: HCl and H₂O are polar (asymmetric charge distribution, net dipole moment); CO₂ and H₂ are non-polar (symmetric, zero net dipole moment). Option 2: electric flux Φ=EA≈1.57Nm2C−1.
Option 1 — Polar vs non-polar molecules
A molecule is polar if its centres of positive and negative charge do not coincide, giving it a permanent (net) dipole moment; it is non-polar if they coincide (net dipole moment zero), usually due to symmetry.
CO₂ — linear, symmetric (O=C=O); the two C=O bond dipoles are equal and opposite, so they cancel ⇒non-polar.
HCl — different atoms (H, Cl) with unequal electronegativity, bond dipole does not cancel ⇒polar.
H₂ — homonuclear diatomic (identical atoms), no charge asymmetry ⇒non-polar.
H₂O — bent (angular) shape; the two O–H bond dipoles do not cancel ⇒polar.
AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marks
Q.A closed spherical surface encloses a charge q at its centre. Show that electric flux through the closed surface is q/ε0.
OR
A pair of charges +q and −q, separated by a small distance 2a is placed in an electric field E, so that the line joining the charges makes an angle θ with E. Write the expressions for torque τ and also its magnitude |τ|.
›Reveal solutionSolution
Gauss's law shows the flux out of a sphere enclosing a point charge q is q/ε₀, independent of the sphere's radius; and the torque on a dipole in a field is τ = pE sinθ.
Main question: Consider a point charge q at the centre O of a spherical surface of radius r. By Coulomb's law, the magnitude of the electric field at every point on this sphere (all at distance r from q) is the same:
E=4πϵ01r2q
and it points radially outward, i.e. parallel to the outward area vector dA at every point on the sphere (by symmetry). So the flux through the whole closed surface is simply E times the total surface area 4πr2:
ΦE=∮E⋅dA=E×4πr2=4πϵ01r2q×4πr2=ϵ0q
Notice the radius r cancels out completely — the flux depends only on the enclosed charge q, not on the size of the sphere. This is exactly Gauss's law for this special symmetric case.