One number for every closed surface
A point charge sends \(q/\varepsilon_0\) through every sphere around it. The result is far more general than that. Take any closed surface at all, a box, a sphere or a shape like a potato. Put any charges inside it and any outside. The flux out through the surface is set by one number only: the net charge inside it.
The total electric flux out through any closed surface equals the net charge enclosed by the surface divided by \(\varepsilon_0\): \(\oint \vec E\cdot d\vec A\) \(= q_\text{in}/\varepsilon_0\). The field \(\vec E\) in the integral is the total field of all charges, inside and outside.
The closed surface is called a Gaussian surface. It is imagined, not made of anything, and it may cut through space or through matter. The charge it holds, counted with signs, is the enclosed charge \(q_\text{in}\). A net positive charge inside sends more lines out than in, and the flux is positive. A net negative charge inside draws more lines in, and the flux is negative.
Look at the first chip of Fig 3.1. Some lines of the \(+2\,\mu\text{C}\) end on the \(-1\,\mu\text{C}\) and never leave S. The net charge inside is \(+1\,\mu\text{C}\), so the flux is \(10^{-6}/8.85 \times 10^{-12}\) \(\approx 1.13 \times 10^5\,\text{N m}^2/\text{C}\). Every line of the outside charge that enters S also leaves it. That charge bends the field all over S, yet it adds nothing to the flux.
The second chip holds a dipole, \(+q\) and \(-q\). The field on S is far from zero; lines leave S near \(+q\) and come back in near \(-q\). But the outward and inward flux cancel exactly, because the net charge inside is zero. Zero flux never means zero field.
In a region the field is \(\vec E = \alpha x\,\hat i\), with \(\alpha = 100\,\text{N/(C m)}\). A cube of side \(a = 0.1\,\text{m}\) has its faces along the axes and runs from \(x = a\) to \(x = 2a\). Find the net flux out of the cube and the charge inside it.
The field points along \(x\), so only the two faces square to the \(x\) axis carry flux. The other four faces have the field running along them.
At \(x = a\): \(\Phi_1\) \(= -(\alpha a)a^2\) \(= -\alpha a^3\). At \(x = 2a\): \(\Phi_2\) \(= +(2\alpha a)a^2\) \(= +2\alpha a^3\).
Net flux \(= -\alpha a^3 + 2\alpha a^3\) \(= \alpha a^3\) \(= 100 \times 10^{-3}\) \(= 0.1\,\text{N m}^2/\text{C}\). By Gauss's law, \(q_\text{in}\) \(= \varepsilon_0 \times 0.1\) \(= 8.85 \times 10^{-13}\,\text{C}\).
More field leaves than enters, so the cube holds a small positive charge. A field that grows along its own direction always means charge is present.
The net flux out of a closed box is \(-3000\,\text{N m}^2/\text{C}\). What is the net charge inside the box, in nC? Take \(\varepsilon_0\) \(= 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)\).
A closed surface holds \(+2\,\mu\text{C}\) and \(-1\,\mu\text{C}\). A charge of \(+5\,\mu\text{C}\) is brought up just outside it. What happens?
Why the law holds
Take a point charge \(q\) and a small patch \(dA\) of a closed surface, at a distance \(r\), as in Fig 3.2. The field at the patch points straight away from the charge. Let the patch's outward normal make an angle \(\theta\) with it. The flux through the patch is the field times the patch's shadow across the field.
\[ d\Phi = E\,dA\cos\theta = \dfrac{q}{4\pi\varepsilon_0 r^2}\,dA\cos\theta \]
Why. the field of a point charge at distance \(r\), times the shadow of the patch across the field.
\[ d\Phi = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{dA\cos\theta}{r^2} = \dfrac{q}{4\pi\varepsilon_0}\,d\Omega \]
Why. \(dA\cos\theta/r^2\) is the solid angle the patch holds at the charge. The \(r^2\) of the field has cancelled the \(r^2\) of the solid angle.
\[ \Phi = \dfrac{q}{4\pi\varepsilon_0}\oint d\Omega = \dfrac{q}{4\pi\varepsilon_0} \times 4\pi = \dfrac{q}{\varepsilon_0} \]
Read it. the patches of a closed surface around the charge hold every direction once, so their solid angles add to \(4\pi\).
The cancellation in step 2 is the heart of the law. It needs a field that falls exactly as \(1/r^2\). The shape of the surface has dropped out: a near patch and a far patch that fill the same thin cone take the same flux. That is why a box, a sphere or a potato around the charge all give \(q/\varepsilon_0\).
Now put the charge outside the surface. Every thin cone from the charge that meets the surface cuts it twice, once going in and once coming out. The two patches hold the same solid angle, and one flux is negative, the other positive. They cancel, so an outside charge gives zero flux. For many charges, the fields add, so the fluxes add. Only the charges inside contribute, and the total is \(q_\text{in}/\varepsilon_0\).
Gravity also falls as \(1/r^2\), so it has a Gauss's law of its own: \(\oint \vec g\cdot d\vec A\) \(= -4\pi G M_\text{in}\). The minus sign says the field points inward, towards mass. The shell theorems of the chapter on Gravitational Force and Field follow from it in two lines. If a force fell as \(1/r^{2.01}\), no such law would hold, and the flux would change with the size of the surface.
Choosing the surface
Gauss's law is true for every closed surface. To find a field from it, we need a surface on which the integral becomes simple. On the parts that carry flux, the field must have one size and cross along the normal. Then the integral is just \(E\) times that area. On every other part, the field must run along the surface and carry no flux.
Only symmetry can promise this before we know the field. A point charge or a charged sphere has spherical symmetry, so the surface is a sphere with the same centre. A long straight line of charge needs a cylinder around it. A large flat sheet needs a small closed cylinder crossing it, called a pillbox. The next two lessons use all three.
You want the field near a long straight wire carrying a uniform charge. Which Gaussian surface lets Gauss's law give the field?
A point charge moves about inside a closed surface, coming near one side and then another. What happens to the total flux out through the surface?
Play the figure above. The charge of \(53.1\,\text{nC}\) moves inside a Gaussian sphere of radius \(10\,\text{cm}\). The field at the top point A changes between about \(2.4 \times 10^4\) and \(9.0 \times 10^4\,\text{N/C}\), and the arrows round the sphere swing and change length. The flux stays \(6000\,\text{N m}^2/\text{C}\) throughout, because 12 lines leave and none returns.
A point charge sits at the centre of a Gaussian sphere. It is moved to another point, still inside the sphere. Which statements are true?
Carl Friedrich Gauss worked on the Earth's magnetism with the physicist Wilhelm Weber in Göttingen. In 1835 he wrote down the law of this lesson for electric charges, as part of that work. He did not publish it. The note appeared in 1867, in his collected works, twelve years after his death. Lagrange had found the same result for gravity in 1773. Gauss's name stayed on the law because his mathematics of surfaces made it general.
- Counting a charge outside the Gaussian surface in \(q_\text{in}\).Only charges inside count. An outside charge changes the field on the surface, not the flux.
- Concluding that zero flux means zero field on the surface.A dipole inside gives zero flux with a strong field on the surface. Zero flux means as much field enters as leaves.
- Adding the flux of two opposite faces with the same sign.Each face has its own outward normal; a field entering one face and leaving the other gives fluxes of opposite sign.