Electric Field Due to an Infinite Sheet of Charge: MDCAT Physics notes

Electric flux

Electric flux is the number of electric field lines passing through a surface:

$$\Phi_E = \vec{E}\cdot\vec{A} = EA\cos\theta$$

$\theta$ is the angle between $\vec{E}$ and the vector area (the normal to the surface).

• $\theta = 0^\circ$: field perpendicular to the surface, flux maximum, $\Phi = EA$.
• $\theta = 90^\circ$: field parallel to the surface, flux zero (minimum).

Unit: N m$^2$ C$^{-1}$. Flux is a scalar.

Gauss's law

The total electric flux through any closed surface equals $1/\varepsilon_0$ times the charge enclosed:

$$\Phi_E = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$$

• The flux depends only on the magnitude of the charge enclosed.
• It does not depend on the size, shape or radius of the closed surface, or on where inside the charge sits.
• Charges outside the surface contribute zero net flux.

Gauss's law is used to find the field of symmetric charge distributions (sphere, sheet, line). It can also be used to find the enclosed charge or flux, but it needs the permittivity to be known already; it cannot determine permittivity.

Field of an infinite sheet of charge

Let a large flat sheet carry surface charge density $\sigma = Q/A$ (C m$^{-2}$). Take a cylindrical Gaussian surface ("pill box") crossing the sheet. The field is perpendicular to the sheet and passes out through both flat ends, each of area $A$:

$$E(2A) = \frac{\sigma A}{\varepsilon_0} \quad\Rightarrow\quad E = \frac{\sigma}{2\varepsilon_0}$$

• $E$ is independent of distance from the sheet: the field is uniform.
• $E \propto \sigma$: a 20% rise in $\sigma$ raises $E$ by 20%.

Field between two oppositely charged plates

For two parallel plates with charge densities $+\sigma$ and $-\sigma$, the fields of the two sheets add between the plates and cancel outside:

$$E = \frac{\sigma}{\varepsilon_0}\ \text{(between)}, \qquad E = 0\ \text{(outside)}$$

In the middle region the field is uniform; it is non-uniform only near the edges (fringing). In terms of potential difference, $E = V/d$. With $V$ kept constant (plates connected to a battery), halving $d$ doubles $E$. If the plates are isolated so that $\sigma$ is fixed, $E = \sigma/\varepsilon_0$ does not change with $d$.

Conductors

In electrostatic equilibrium the net charge of a conductor lies on its outer surface. The charge inside, and the field inside, are zero.

Key formulas

• $\Phi_E = EA\cos\theta$
• $\Phi_E = Q/\varepsilon_0$
• Sheet: $E = \sigma/2\varepsilon_0$; plates: $E = \sigma/\varepsilon_0 = V/d$

Common MDCAT traps

• Single sheet is $\sigma/2\varepsilon_0$; two opposite plates give $\sigma/\varepsilon_0$.
• Flux is zero at $90^\circ$ between $E$ and the area vector, maximum at $0^\circ$.
• Flux through a sphere depends on the enclosed charge, not its radius or area.
• The field between plates is uniform, not variable.
• Gauss's law cannot give permittivity.

Quick revision

• Field of an infinite sheet does not fall off with distance.
• $E \propto \sigma$.
• Charge on a conductor resides on its outer surface.
• Flux is a scalar; unit N m$^2$ C$^{-1}$.
• Gauss's law applies to any closed surface.