Electric Field Lines and Infinite Sheet Field
The Area Vector and Electric Flux
Electric Flux ($\Phi_e$) quantifies the total electric field 'flowing' through a specific surface, depending on field strength, area, and orientation.
$$\Phi_e = \vec{E} \cdot \vec{A} = EA\cos\theta$$
Flux is the scalar product of the electric field vector and the area vector.
$\Phi_e$=Electric Flux(Nm²/C)
$E$=Electric Field magnitude(N/C)
$A$=Area of the surface(m²)
$\theta$=Angle between the field lines and the normal to the surface(degrees or radians)
$\theta = 0°$
→Field perpendicular to surface (parallel to normal): $\Phi = EA$ (maximum)
$\theta = 90°$
→Field parallel to surface (perpendicular to normal): $\Phi = 0$
$\theta = 180°$
→Field anti-parallel to normal: $\Phi = -EA$ (maximum negative — flux entering)
Area Vector: Represented by $\vec{A}$ pointing perpendicular (normal) to the surface. For closed surfaces, the normal always points outward.
Scalar Quantity: Flux is a dot product, so it is a scalar — it can be positive (leaving), negative (entering), or zero.
Dimensional Check: $[\Phi] = [E][A] = \text{N/C} \times \text{m}^2 = \text{Nm}^2\text{/C}$ — always verify your answer has these units.
Flux Through a Closed Surface
For a point charge $q$ at the center of a sphere of radius $r$, the total flux through the sphere can be computed by summing contributions from tiny flat patches — leading to the key result $\Phi_e = q/\epsilon_0$.
$$\Phi_e = E \times 4\pi r^2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2} \times 4\pi r^2 = \frac{q}{\epsilon_0}$$
The $r^2$ in the denominator of Coulomb's field cancels the $r^2$ in the sphere's surface area, making flux independent of radius.
$E$=Field magnitude at the sphere surface(N/C)
$4\pi r^2$=Total surface area of the sphere(m²)
Surface is not a sphere
→Same result — total flux through ANY closed surface enclosing $q$ is $q/\epsilon_0$, regardless of shape
Why Spherical Symmetry Works: On a sphere centered on $q$, the field is uniform in magnitude and everywhere perpendicular to the surface, so $\cos\theta = 1$ at every patch.
Shape Independence: The total flux depends only on the enclosed charge and $\epsilon_0$ — not on the shape or size of the closed surface.
Physical Insight: Every field line originating from $q$ must pierce any surrounding closed surface exactly once, no matter how distorted.
Gauss's Law
Gauss's Law relates the total Electric Flux through a closed Gaussian Surface to the net charge enclosed within it.
$$\Phi_e = \frac{Q_{enclosed}}{\epsilon_0}$$
The net flux through any closed surface equals the total enclosed charge divided by the permittivity of free space.
$\Phi_e$=Net electric flux through the closed surface(Nm²/C)
$Q_{enclosed}$=Algebraic sum of all charges inside the surface(C)
$\epsilon_0$=Permittivity of free space (≈ 8.85 × 10⁻¹² C²/Nm²)(C²/Nm²)
$Q_{enclosed} = 0$
→Net flux is zero — but $E$ need not be zero on the surface (external fields still exist)
$Q_{enclosed} > 0$
→Net flux is positive (more lines leaving than entering)
$Q_{enclosed} < 0$
→Net flux is negative (more lines entering than leaving)
Geometry Independence: Flux depends only on the charge enclosed, not the shape or size of the surface.
External Charges Ignored: Charges outside the surface contribute zero net flux — every field line entering must also exit.
Algebraic Sum: $Q_{enclosed} = q_1 + q_2 + \ldots + q_n$ — signs matter. A dipole ($+q$ and $-q$) inside gives zero net flux.
Applications of Gauss's Law
Inside a hollow charged conducting sphere, the electric field is exactly zero — a direct consequence of Gauss's Law leading to Electrostatic Shielding.
$$E_{inside} = 0 \quad (r < R)$$
A Gaussian sphere of radius $r < R$ encloses zero charge (all charge resides on the outer surface of a conductor), so $\Phi = 0$ and hence $E = 0$.
$R$=Radius of the conducting sphere(m)
$r$=Distance from center (observation point)(m)
$r = R$ (at the surface)
→$E = \frac{1}{4\pi\epsilon_0}\frac{q}{R^2}$
$r > R$ (outside)
→Field behaves as if all charge $q$ were a point charge at the center: $E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$
Why Zero Inside: In a conductor at equilibrium, charges repel to the outer surface. A Gaussian surface inside encloses no charge → $\Phi = 0$ → $E = 0$.
Electrostatic Shielding: Sensitive electronics (TVs, computers) are enclosed in metal boxes (Faraday cages) to block external electric fields.
Outside the Sphere: For $r > R$, the field is identical to that of a point charge $q$ at the center — the shell theorem for electrostatics.
An infinite plane sheet of charge with uniform surface charge density $\sigma$ produces a uniform electric field that is independent of distance from the sheet.
$$E = \frac{\sigma}{2\epsilon_0}$$
Using a cylindrical Gaussian surface piercing the sheet, flux exits through both flat end faces. The curved surface contributes nothing because the field is parallel to it.
$\sigma$=Surface charge density (charge per unit area)(C/m²)
$\epsilon_0$=Permittivity of free space(C²/Nm²)
Distance from sheet → ∞
→Field is still $\sigma/2\epsilon_0$ — the infinite extent means it never falls off (an idealization)
Gaussian Surface Choice: A cylinder with flat faces on either side of the sheet. Field exits both faces: $2EA = \sigma A/\epsilon_0$, giving $E = \sigma/2\epsilon_0$.
No Distance Dependence: Unlike point charges ($E \propto 1/r^2$), the field of an infinite sheet is constant everywhere — a key conceptual difference.
Direction: Field points away from the sheet on both sides (for positive $\sigma$).
Between two oppositely charged infinite parallel plates ($+\sigma$ and $-\sigma$), the individual fields add by superposition between the plates and cancel outside.
$$E_{between} = \frac{\sigma}{\epsilon_0}$$
Each plate contributes $\sigma/2\epsilon_0$. Between the plates both fields point the same direction (from + to −), so they add. Outside, they cancel.
$\sigma$=Surface charge density on each plate ($q/A$)(C/m²)
$\epsilon_0$=Permittivity of free space(C²/Nm²)
Outside the plates
→Fields from the two plates point in opposite directions and cancel: $E_{outside} = 0$
Superposition: Between plates, both individual fields point from + to −, so $E = \sigma/2\epsilon_0 + \sigma/2\epsilon_0 = \sigma/\epsilon_0$.
Uniform Field: The field is constant at all points between the plates — the basis of parallel-plate capacitors.
Gaussian Box Method: A box with one face inside the metal plate ($E = 0$ there) and one in the gap gives $EA = \sigma A/\epsilon_0$ directly.