Electric flux

What Flux Means Physically

Flux describes the amount of a given quantity that passes through a given area. Think of it like counting how many field lines pierce a surface.

A large flux means many field lines pass through the surface. A small flux means few do. If field lines run parallel to the surface, none pierce it, so the flux is zero. The concept connects the geometry of a surface to the field passing through it.

Flux depends on three things: the field strength, the area of the surface, and the angle between the field and the surface. A surface facing the field head-on captures maximum flux. A surface tilted away captures less. A surface edge-on to the field captures none.

MisconceptionFlux is not the same as electric field strength. A strong field through a tiny area can produce less flux than a weak field through a huge area.

Exam TipAlways consider both field magnitude and area when comparing fluxes.

The Area Vector

Before computing flux, understand the area vector A⃗. Every flat surface has a vector associated with it.

The area vector has a magnitude equal to the surface area $A$. Its direction is perpendicular to the surface. For a closed surface [a surface that fully encloses a volume, like a sphere or a cube], the area vector always points outward. For an open surface, either perpendicular direction may be chosen, but stay consistent.

Reading the notation: when you see $\vec{A}$ or $d\vec{A}$, both carry directional information. The lowercase $d\vec{A}$ represents an infinitesimal patch of area, used when the surface is curved or the field varies across it.

Exam TipThe angle θ is always measured between E⃗ and A⃗, not between E⃗ and the surface itself.

Electric Flux Through a Uniform Field

For an electric field E⃗ that is constant across an area A⃗, the electric flux is given by the dot product of the two vectors.

Key Equations

Electric flux (uniform field):

$${\Phi }_{E}=\vec{E}\cdot \vec{A}=EAcos\theta $$

Variables:

• ${\Phi }_{E}$ = electric flux (N·m²/C)
• $E$ = electric field magnitude (N/C)
• $A$ = area of the surface (m²)
• $\theta $ = angle between E⃗ and A⃗ (rad or °)

SI unit: N·m²/C (equivalently V·m)

ProportionalityFlux is directly proportional to both $E$ and $A$. Doubling the field strength doubles the flux. Doubling the area doubles the flux.

Reference sheet status: On the reference sheet.

The dot product determines the sign of flux. When $\theta <90^{\circ}$, the dot product is positive, so flux is positive. Field lines exit the surface. When $\theta >90^{\circ}$, the dot product is negative, so flux is negative. Field lines enter the surface. When $\theta =90^{\circ}$, the dot product is zero. No field lines pierce the surface.

For a closed surface, positive flux means net field lines leaving the enclosed volume. Negative flux means net field lines entering.

Examiner InsightMany students forget to use the angle between E⃗ and A⃗. They mistakenly use the angle between E⃗ and the surface plane, giving the complementary angle.

Exam TipIf given the tilt angle of the surface from the field, subtract it from 90° to get θ.

Worked Example: Flux Through a Tilted Surface

Scenario

A uniform electric field of magnitude 500 N/C points in the +x-direction. A flat square surface of side length 0.30 m is oriented so that its area vector makes an angle of 60° with the electric field.

Step-by-step solution:

Equation used

$${\Phi }_{E}=EAcos\theta $$

Given

$$E=500\text{ N/C}$$

$$A=(0.30{)}^{2}=0.090{\text{ m}}^{2}$$

$$\theta =60^{\circ}$$

Working

$${\Phi }_{E}=(500)(0.090)cos60^{\circ}$$

$${\Phi }_{E}=(500)(0.090)(0.50)$$

$${\Phi }_{E}=22.5{\text{ N}\cdot\text{m}}^{2}\text{/C}$$

Answer

$${\Phi }_{E}=22.5{\text{ N}\cdot\text{m}}^{2}\text{/C}$$

Interpretation

The positive flux means field lines pass outward through the surface. Only half the maximum possible flux passes through because the surface is tilted 60° from the field direction.

Total Flux Through an Arbitrary Surface

When the electric field varies across a surface, or the surface is curved, a single value of $Ecos\theta $ cannot represent the entire surface. The total electric flux is then defined by a surface integral.

Key Equations

Electric flux (general, surface integral):

$${\Phi }_{E}=\oint \vec{E}\cdot d\vec{A}$$

Variables:

• ${\Phi }_{E}$ = total electric flux through the surface (N·m²/C)
• $\vec{E}$ = electric field at each infinitesimal patch (N/C)
• $d\vec{A}$ = infinitesimal area vector, magnitude $dA$, direction outward-normal (m²)

SI unit: N·m²/C

Calculus form: This IS the calculus form. The circle on the integral sign (∮) indicates integration over a closed surface.

Reference sheet status: On the reference sheet.

Reading the notation: the symbol $\oint $ means “integrate over a closed surface.” The integral sums up $\vec{E}\cdot d\vec{A}$ over every tiny patch $d\vec{A}$ of the entire surface. Each patch has its own local direction and its own local field value.

The procedure for evaluating this integral has three parts. First, break the closed surface into sections where the field behavior is uniform. Second, on each section, determine whether $\vec{E}$ is parallel to $d\vec{A}$, perpendicular to $d\vec{A}$, or zero. Third, evaluate each section’s contribution and sum them.

On sections where E⃗ is perpendicular to $d\vec{A}$, the dot product is zero. Those sections contribute no flux. On sections where E⃗ is parallel to $d\vec{A}$ and constant in magnitude, $\vec{E}\cdot d\vec{A}=E dA$. The integral simplifies to $E$ times the area of that section. This simplification is the key to applying Gauss’s law, covered in the next topic.

For a closed surface in a uniform field with no enclosed charge, every field line that enters one side exits the other. The positive and negative contributions cancel, giving zero net flux.

Exam TipFor a closed surface, always draw the area vectors outward on every face before computing the dot product.

MisconceptionStudents sometimes think a closed surface always has nonzero flux because field lines pass through it. If no charge is enclosed, every line that enters also exits, so the net flux is zero.

Exam TipNet flux through a closed surface depends on enclosed charge, not on external fields.

Worked Example: Net Flux Through a Cube in a Uniform Field

Scenario

A cube of side length 0.20 m sits in a uniform electric field $E=400$ N/C directed in the +x-direction. No charge is inside the cube. Calculate the flux through the right face, the left face, and the total flux through the cube.

Step-by-step solution:

Area of one face:

$$A=(0.20{)}^{2}=0.040{\text{ m}}^{2}$$

Right face: The outward area vector points in the +x-direction, parallel to E⃗. So $\theta =0^{\circ}$.

$${\Phi }_{\text{right}}=EAcos0^{\circ}=(400)(0.040)(1)=16.0{\text{ N}\cdot\text{m}}^{2}\text{/C}$$

Left face: The outward area vector points in the −x-direction, antiparallel to E⃗. So $\theta =180^{\circ}$.

$${\Phi }_{\text{left}}=EAcos180^{\circ}=(400)(0.040)(-1)=-16.0{\text{ N}\cdot\text{m}}^{2}\text{/C}$$

Top, bottom, front, and back faces: The outward area vectors are perpendicular to E⃗. So $\theta =90^{\circ}$ and each contributes zero flux.

Total flux:

$${\Phi }_{\text{total}}=16.0+(-16.0)+0+0+0+0=0{\text{ N}\cdot\text{m}}^{2}\text{/C}$$

Answer

$${\Phi }_{\text{total}}=0{\text{ N}\cdot\text{m}}^{2}\text{/C}$$

Interpretation

The net flux is zero because no charge is enclosed. Every field line entering the left face exits the right face.