Electric Flux

The idea

The definition is modelled on the flow of a liquid. If a liquid of velocity $\vec{v}$ crosses a small flat surface $dS$ normally, the volume crossing per unit time is $v\,dS$; if the normal to the surface makes an angle $\theta$ with $\vec{v}$, the projected area is $dS\cos\theta$ and the flux is $\vec{v}\cdot\hat{n}\,dS$. The electric flux is the exact analogue for $\vec{E}$ — but note that nothing physically flows in the electric case.

Area as a vector

• The orientation of an area matters, not just its size: how much water passes through a ring held in a stream depends on how you hold it. So an area element is treated as a vector.
• The direction of a planar area vector is along its normal: $\Delta\vec{S} = \Delta S\,\hat{n}$.
• A curved surface is divided into many small elements, each treated as planar, each with its own $\hat{n}$.
• A normal can point either way. For a closed surface, the convention resolves the ambiguity: $\hat{n}$ is always the outward normal.

Definition

$$\Delta\phi = \vec{E}\cdot\Delta\vec{S} = E\,\Delta S\cos\theta$$

• $\theta$ is the angle between $\vec{E}$ and the area vector (the outward normal for a closed surface), never the angle between $\vec{E}$ and the plane of the surface.
• Read either as $E\,(\Delta S\cos\theta)$ — the field times the area projected normal to $\vec{E}$ — or as $E_\perp\,\Delta S$, the component of $\vec{E}$ along the normal times the area.
• Physically, the electric flux through a surface is the number of electric field lines crossing that surface normally.
• Total flux through a surface $S$: $\phi = \displaystyle\sum\vec{E}\cdot\Delta\vec{S}$, which in the limit $\Delta S\to0$ becomes $\phi = \displaystyle\int_S\vec{E}\cdot d\vec{S}$, and $\displaystyle\oint_S\vec{E}\cdot d\vec{S}$ for a closed surface.

Nature and unit

• Electric flux is a scalar quantity — quote a sign, never a direction. Positive flux means field lines leaving the surface, negative flux means lines entering it.
• SI unit: $\text{N m}^2\,\text{C}^{-1}$, equivalently $\text{V m}$. Dimensions $[\text{ML}^3\text{T}^{-3}\text{A}^{-1}]$.

Special cases worth memorising

• Surface held normal to $\vec{E}$ ($\theta = 0^\circ$): $\phi = ES$, the maximum.
• Surface held with its plane containing $\vec{E}$ ($\theta = 90^\circ$): $\phi = 0$. Field lines graze the surface and cross nothing.
• Curved surface of a cylinder whose axis is along $\vec{E}$: $\phi = 0$, since $\vec{E}$ is everywhere perpendicular to the outward normal there.
• Closed surface in a uniform field: net flux zero — the flux entering one face leaves the other. A cylinder of radius $r$ and length $l$ placed with its axis parallel to a uniform field has $\phi_A = +E\pi r^2$, $\phi_B = -E\pi r^2$, $\phi_{\text{curved}} = 0$, total zero.

Worked patterns CBSE sets

• Square of side $10\,\text{cm}$ held normal to $\vec{E} = 3\times10^{3}\hat{i}\ \text{N C}^{-1}$. Area $\Delta\vec{S} = 10^{-2}\hat{i}\ \text{m}^2$, so $\phi = (3\times10^{3}\hat{i})\cdot(10^{-2}\hat{i}) = 30\ \text{N m}^2\,\text{C}^{-1}$.
• Same square with its plane parallel to the $y$-$z$ plane in $\vec{E} = 5\times10^{3}\hat{i}\ \text{N C}^{-1}$: the area vector is along $\hat{i}$, so $\phi = 5\times10^{3}\times0.01 = 50\ \text{N m}^2\,\text{C}^{-1}$. If the plane is then turned to make $30^\circ$ with the $x$-axis, the normal makes $60^\circ$ with $\vec{E}$, and $\phi' = ES\cos60^\circ = 25\ \text{N m}^2\,\text{C}^{-1}$.
• Square sheet of side $a$ lying parallel to the $XY$ plane at $z = a$, in $\vec{E} = cz^2\hat{k}$. The area vector is $a^2\hat{k}$, so $\phi = cz^2a^2$, and at $z=a$ this is $a^4c$.
• Square plane sheet of side $10\,\text{cm}$ inclined at $30^\circ$ to a uniform field of $200\ \text{N C}^{-1}$: the normal is at $60^\circ$, so $\phi = 200\times0.01\times\cos60^\circ = 1.0\ \text{N m}^2\,\text{C}^{-1}$.
• Cylinder of length $20\,\text{cm}$ and radius $5\,\text{cm}$, axis along $x$, centred at the origin, in a field $\vec{E} = 100\hat{i}\ \text{N C}^{-1}$ for $x>0$ and $-100\hat{i}\ \text{N C}^{-1}$ for $x<0$. At each flat face the field points along the outward normal, so each contributes $+100\,\Delta S$, the curved surface contributes nothing, and $\phi = 200\times\pi(0.05)^2 = 1.57\ \text{N m}^2\,\text{C}^{-1}$.

Reading the angle correctly

Two phrasings appear and they mean different things.