Electric flux

Electric flux is the closing idea of CBSE Class 12 Physics, Unit I (Electrostatics), chapter "Electric Charges and Fields" — it is squarely inside the 2025-26 prescribed scope. The chapter builds up from charge and Coulomb's law, through the electric field and field lines, to the dipole, and finishes here: flux, the statement of Gauss's theorem, and its use for three named symmetric cases. You are expected to define flux, compute it for a flat surface in a uniform field at various angles, state Gauss's theorem, and apply it (never derive it from Coulomb's law — that derivation is explicitly out of scope) to an infinite line charge, an infinite plane sheet, and a uniformly charged thin spherical shell. Fields of a charged solid sphere, a finite cylinder, or any non-symmetric shape are not examinable here.

Picture the electric field as a set of arrows filling space, densest where the field is strong. Electric flux through a small flat patch asks a simple question: how much of that arrow-field is aimed straight through the patch, rather than skimming along its surface? For a uniform field $\vec E$ crossing a flat surface of area $A$, whose orientation is captured by the outward unit normal $\hat n$, the flux is

where: $\vec A = A\hat n$ is the "area vector" — a vector whose magnitude is the surface's area and whose direction is along the outward normal — and $\theta$ is the angle between $\vec E$ and $\hat n$. Physically, $E\cos\theta$ is the component of the field actually aligned with the normal, so flux is that useful component multiplied by the area it acts over. When the field points straight along the normal ($\theta = 0°$), every field line crosses the surface and $\phi = EA$ is as large as it can be. When the field lies flat in the plane of the surface ($\theta = 90°$), no line crosses it at all, and $\phi = 0$, however strong $E$ is. For $\theta$ between $90°$ and $180°$, $\cos\theta$ turns negative, meaning field lines are leaving through that face rather than entering it — flux carries a sign, not just a size. For a surface that is curved, or a field that varies from point to point, the same idea is applied to infinitesimal patches $d\vec A$ and summed:

where: the small circle on the integral sign marks a closed surface (one that fully encloses a volume, like a sphere or a cube, as opposed to an open sheet), and the sum runs over every tiny patch of that surface, each contributing $\vec E\cdot d\vec A$. CBSE does not ask you to evaluate this integral from first principles except in the three symmetric cases below, where a clever choice of surface makes the sum collapse to simple multiplication. The SI unit of flux is $\text{N}\,\cdot\,\text{m}^2/\text{C}$ (equivalently $\text{V}\,\cdot\,\text{m}$), and it is a pure scalar — direction has already been folded into the dot product, so there is nothing left to point.

Gauss's theorem (statement only). The total electric flux through any closed surface (a "Gaussian surface") equals $1/\varepsilon_0$ times the total charge enclosed within it:

where: $q_{\text{enclosed}}$ is the algebraic sum of all charge sealed inside the surface (charges outside contribute nothing to the total flux, even though they still contribute to $\vec E$ at points on the surface), and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space. This is a completely general result — true for any closed surface, any shape, any charge arrangement — but CBSE asks you to state it, not derive it from Coulomb's law. It becomes a practical calculation tool only when you can choose a Gaussian surface over which $\vec E$ is either constant in magnitude and parallel to $d\vec A$ everywhere (so the integral becomes $E\times\text{area}$), or exactly perpendicular to $d\vec A$ (contributing zero). That freedom to choose the surface is the entire trick behind the three standard applications below.

• Infinite line charge, linear charge density $\lambda$ (charge per unit length): choosing a coaxial cylindrical Gaussian surface of radius $r$ and length $l$, the flux through the curved part alone is $E\times 2\pi r l$ (the flat end-caps contribute zero, since $\vec E$ there runs parallel to the caps, not through them), and Gauss's theorem gives $E\times 2\pi r l = \lambda l/\varepsilon_0$, so

directed radially outward for positive $\lambda$, falling off as $1/r$.

• Infinite plane sheet, uniform surface charge density $\sigma$ (charge per unit area): choosing a pillbox-shaped Gaussian surface straddling the sheet, with two flat faces of area $A$ parallel to the sheet, symmetry gives equal outward flux $EA$ through each face, so $2EA = \sigma A/\varepsilon_0$, hence

uniform in magnitude and direction, independent of distance from the sheet — one of the most counter-intuitive and most-tested results in the chapter.

• Uniformly charged thin spherical shell, total charge $q$, radius $R$: for a point outside the shell ($r>R$), a concentric spherical Gaussian surface of radius $r$ encloses all of $q$, giving $E\times4\pi r^2 = q/\varepsilon_0$, so $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ — identical to a point charge $q$ sitting at the centre; for a point inside the shell ($r<R$), a Gaussian sphere of that radius encloses zero charge (all the charge sits on the shell, outside this surface), so $E=0$ everywhere inside.

Board level - Flux through a tilted flat surface

Question: A uniform electric field $\vec E = 500\ \text{N/C}$ passes through a flat square surface of side $0.20\ \text{m}$. Find the electric flux through the surface when (a) the field is perpendicular to the surface, and (b) the field makes an angle of $60°$ with the surface normal.

Step 1. Compute the area of the flat square surface: $A = (0.20\ \text{m})^2 = 0.04\ \text{m}^2$.

Step 2. Case (a): the field is perpendicular to the surface, so $\theta = 0°$ and $\cos\theta = 1$. Then $\phi = EA\cos\theta = 500\times0.04\times1 = 20\ \text{N}\,\cdot\,\text{m}^2/\text{C}$.

Step 3. Case (b): the field makes $\theta = 60°$ with the normal, and $\cos60° = 0.5$. Then $\phi = EA\cos\theta = 500\times0.04\times0.5 = 10\ \text{N}\,\cdot\,\text{m}^2/\text{C}$.

Step 4. Compare: tilting the surface by $60°$ away from face-on halved the flux, exactly tracking $\cos60°=0.5$, even though the field strength and the surface area never changed.

Answer: Flux is $\mathbf{20\ N\cdot m^2/C}$ when perpendicular and $\mathbf{10\ N\cdot m^2/C}$ at $60°$ — flux falls off as $\cos\theta$ purely because of orientation.

Competitive level - Total and per-face flux through a cube from an enclosed point charge

Question: A point charge $q = +8.85\times10^{-8}\ \text{C}$ is placed at the centre of a cube of side $a$. Find the total electric flux through the cube's closed surface, and the flux through one of its six faces. Then find the flux through one face if the same charge is instead placed at one corner of a cube of the same size, shared among eight such cubes meeting at that corner.

Step 1. Apply Gauss's theorem to the whole closed cube directly: $\phi_{\text{total}} = q_{\text{enclosed}}/\varepsilon_0$, and this holds regardless of the cube's shape — only the enclosed charge matters, not the surface's geometry.

Step 2. Substitute the numbers: $\phi_{\text{total}} = \dfrac{8.85\times10^{-8}}{8.85\times10^{-12}} = 1\times10^{4}\ \text{N}\,\cdot\,\text{m}^2/\text{C}$.

Step 3. Since the charge sits exactly at the cube's centre, all six faces are geometrically equivalent by symmetry, so each face intercepts an equal one-sixth share: flux through one face $= \dfrac{1\times10^4}{6}\approx1.67\times10^3\ \text{N}\,\cdot\,\text{m}^2/\text{C}$.

Step 4. For the corner case, imagine the same point charge sitting at the shared corner of eight identical cubes stacked around it (two along each axis) — by the same symmetry argument, the total flux $q/\varepsilon_0$ splits equally among the eight cubes, so each cube gets $\phi_{\text{cube}} = \dfrac{1}{8}\times1\times10^4 = 1.25\times10^3\ \text{N}\,\cdot\,\text{m}^2/\text{C}$, and since each cube contributes three faces meeting at that corner (with the other three faces getting zero flux by symmetry of the corner geometry), flux through one of those three faces is $\dfrac{1.25\times10^3}{3}\approx4.17\times10^2\ \text{N}\,\cdot\,\text{m}^2/\text{C}$.

Answer: Centred charge: total flux $\mathbf{1\times10^4\ N\cdot m^2/C}$, flux per face $\mathbf{\approx1.67\times10^3\ N\cdot m^2/C}$. Corner charge shared by eight cubes: flux per cube $\mathbf{1.25\times10^3\ N\cdot m^2/C}$. Timing note: In a competitive exam, resist the urge to integrate anything here — the moment you see "point charge enclosed by a closed surface," reach straight for $q/\varepsilon_0$, then use symmetry (equal faces, equal cubes) to split it, which takes under a minute; only fall back to $\oint\vec E\cdot d\vec A$ by brute force if the geometry genuinely lacks a symmetry to exploit.

• Forgetting the $\cos\theta$ factor and computing flux as simply $EA$ regardless of the angle between the field and the surface normal — this alone accounts for a large share of lost marks in flux-calculation questions.
• Believing charges outside a closed surface contribute nothing to $\vec E$ at points on that surface — they do contribute to the local field, but their net contribution to the total flux through the closed surface is exactly zero, because every field line from an outside charge that enters the surface must also exit it somewhere else.
• Trying to use Gauss's theorem to compute $E$ for shapes that lack the needed symmetry (for example, a finite line segment or an irregular blob of charge) — the theorem remains true, but it stops being useful because $E$ cannot be pulled out of the surface integral as a constant.
• Confusing linear charge density $\lambda$ (units C/m), surface charge density $\sigma$ (units C/m²), and total charge $q$ (units C) inside the line-charge and sheet-charge formulas — mixing these up gives an answer with the wrong power of length.
• Treating the sheet-charge result $E=\sigma/2\varepsilon_0$ as though it weakens with distance, when in fact it is famously constant — a common source of "the field must fall off like $1/r^2$" reflex errors carried over from the point-charge formula.
• Getting the sign of flux backwards: flux is positive where field lines leave a closed surface and negative where they enter, so a closed surface with equal charge entering and leaving nets to zero total flux even though the field is nonzero everywhere on it.

• Define electric flux and compute it for a flat surface in a uniform field, including cases where the field is tilted at an angle to the surface normal.
• Explain, in words and with the cosine factor, why flux is maximum when the field is perpendicular to a surface and zero when the field lies in the surface's plane.
• State Gauss's theorem in symbols and in words, and explain why only the enclosed charge (not external charge, not the charge's exact position inside) fixes the total flux through a closed surface.
• Apply Gauss's theorem, using a symmetry argument to choose the right Gaussian surface, to state the field of an infinite line charge, an infinite charged plane sheet, and a uniformly charged spherical shell (both inside and outside).
• Split a known total flux across identical faces or sub-volumes using symmetry, as in a point charge at the centre or at a corner of a cube.

Ask the tutor: want to walk through why the Gaussian-surface choice differs between a line, a sheet, and a sphere, or need a hand with a trickier flux-splitting problem like the cube-corner case above — ask the AI tutor to work through it step by step.