Electric Flux and Gauss's Law: Quick Revision

A brief note on Electric Flux and Gauss's Law. Read it before a test. Try each question at the end before you open its answer.

Electric flux

• Flat surface in a uniform field: $\Phi = EA\cos\theta = \vec{E}\cdot\vec{A}$. The angle $\theta$ is between the field and the normal to the surface.
• Unit: $\text{N m}^2/\text{C}$, the same as the volt metre. Flux is a scalar.
• Facing the field: $+EA$. Along the field: $0$. Facing away: $-EA$.
• Angle with the plane of the surface: use the sine in place of the cosine.
• Closed surface: the normal points outward. Lines that leave count as positive, lines that enter count as negative.
• Uniform field: the net flux through any closed surface is zero.

Gauss's law

• Statement: $\oint \vec{E}\cdot d\vec{S} = \dfrac{q_{\text{in}}}{\varepsilon_0}$, with $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)$.
• Only the net charge inside counts. The shape and the size of the surface do not matter.
• A charge outside sends as many lines in as out. It changes the field on the surface, but not the net flux.
• The law follows from the inverse square law. It is true for every closed surface, but it gives the field only when the charges have a symmetry.
• Point charge at the centre of a cube: each face gets $\dfrac{q}{6\varepsilon_0}$.

Fields from Gauss's law

• Long wire: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$. It falls as $1/r$.
• Large sheet: $E = \dfrac{\sigma}{2\varepsilon_0}$, the same at every distance. Two opposite sheets give $\dfrac{\sigma}{\varepsilon_0}$ between them and zero outside.
• Thin shell, outside: $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$. Inside: $E = 0$.
• Solid ball, inside: $E = \dfrac{Qr}{4\pi\varepsilon_0 R^3}$. Outside: the same as a point charge.

Conductors

• The field inside a conductor at rest is zero, and the charge sits on the surface.
• Just outside the surface the field is at right angles to it, with $E = \dfrac{\sigma}{\varepsilon_0}$.
• The field in an empty cavity is zero. A charge $+q$ in a cavity draws $-q$ on the wall of the cavity.

A worked example

A point charge of $6.6$ nC is at the centre of a cube. Find the flux through the cube and through one face.

1. The cube. $\Phi = \dfrac{6.6 \times 10^{-9}\ \text{C}}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 746\ \text{N m}^2/\text{C}$.
2. One face. $\dfrac{746}{6} = 124\ \text{N m}^2/\text{C}$. The size of the cube does not matter.

Practise with new numbers

Check yourself

1. A square frame of side $0.30$ m is held so that its plane makes $30^\circ$ with a uniform field of $90$ N/C. What is the flux through it?

2. A point charge sits just outside a closed surface. Is the net flux through the surface zero? Is the field on the surface zero?

3. A large sheet makes a field of $E$ at a point $1$ m away. What field does it make at $3$ m?