Gauss's Law

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Electric flux and charge

Why is the direction of the electric field chosen to be directed outward from positive charges and inward towards negative charges. It turns out that it is a mathematical convenience brought about by Gauss's Law. That is

"for any closed surface that surrounds a volume distribution of charge, the total electric flux passing through that surface must be equivalent to the total enclosed charge."

Another consequence of this is

"all field lines must emanate from or terminate on a charge."

Mathematically, this means

$Q_\textrm{encl} = \oint \textbf{D}\cdot \hat{\textbf{n}} dS$ ,

where $\hat{\textbf{n}}$ is a unit vector that points outward from the surface.

If we try it for a sphere of radius a enclosing a point charge $q 1$ at the origin (whose field is defined by $\textbf{E}$ in the preceding section), we get

$Q_\textrm{encl} = \int^{\pi}_{0}\!\!\int_{0}^{2\pi}\frac{q_1}{4\pi a^2} a^2 \sin\theta\, d\theta\, d\phi$

Simple illustration of Gauss's Law using a point charge at the origin and a spherical enclosing surface.

Carrying out the integration, it is easy to see that $Q encl = q 1$ . The unit vector is not explicitly seen here, but it is easy to verify that the direction of electric field of the point charge coincides perfectly with the radially directed unit vector that points away (outward) from our chosen spherical surface. Because of this choice of conventions, the electric field from positive charges is assumed to point outward (and vice-versa for negative charges). This also makes Gauss's Law consistent with the Divergence Theorem, which will be covered in more detail later.

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An interesting example: the Electric Dipole

Imagine two charges of equal magnitude, but opposite sign spaced by a small distance a. What happens when we apply Gauss's Law to a spherical surface enclosing the two charges? Since we know that the total electric flux passing through the surface must be equivalent to the enclosed electric charge, the surface integral that defines Gauss's Law must be zero. Let us briefly prove it.

Geometry of the point-charge electric dipole. Spherical coordinate system shown.

The electric field of the dipole configuration (if the observation distance R is much larger than the dipole spacing 2b) is

$E_r = \frac{q\, b}{\pi\epsilon_0 R^3}\, \cos\theta$

$E_\theta = - \frac{q\, b}{2\pi\epsilon_0 R^3}\, \sin \theta.$

(The derivation of the dipole field will come a bit later.) Notice the inverse cube variation with distance for the dipole field.

Now, let us apply the Gauss integral over the same sphere of radius a used in the point-charge case. The unit vector $\hat{\textbf{n}} = \hat{\textbf{r}}$ , so only the radial component of the electric field $E_r\,$ contributes and we have (after recognising that $\textbf{D} = \epsilon_0 \textbf{E}$ )

$Q_\textrm{encl} = \frac{q\, b}{\pi\, a} \int_0^{2\pi}\!\!\int_0^\pi \cos\theta \sin\theta\, d\theta\, d\phi.$

Electric field lines for electric dipole in the x-z plane. Field lines point outward at the top (where the positive charge is). The lines curve around to the bottom where they reenter the circle at r=1 where the negative charge is found.

After carrying out the integration, we see that $Q_\textrm{encl}\,$ indeed vanishes. This is because integrating the product of sine and cosine over a half-period yields zero. Intuitively, this means that the number of "field lines" leaving the spherical surface is exactly equal to the "field lines" entering the surface. See the accompanying illustration.

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The Divergence Theorem

Gauss's Law is closely related to the Divergence Theorem of multivariable calculus. In fact, Gauss's law forms part of the definition of divergence, viz.

$\nabla\cdot\textbf{D} = \lim_{\Delta V \rightarrow 0}\frac{\oint_{\partial V} \textbf{D} \cdot \hat{\textbf{n}} \, dS}{\Delta V},$

where $\partial V$ represents a closed surface around the volume $\Delta V\,$ . In fact, we can use a hexahedral volume with rectangular sides to illustrate the definition in more detail.

Rectangular volume enclosed with a closed surface of rectangular faces. The vector field D is not shown, but assumed to be present.

In the figure, consider the two faces "1" and "2" at (x, y, z) and (x, y+dy, z). The normal unit vectors are $-\hat{\textbf{y}}$ and $\hat{\textbf{y}}$ , respectively. At (x, y, z) the flux is given by

$\Phi_1 = -D_y(x, y, z)\, dx\, dz$ .

At "2", we have

$\Phi_2 = D_y(x, y+dy, z) \, dx\, dz$ .

If we take the difference in the fluxes for these two faces and divide by the incremental volume $dV = dx\, dy\, dz\,$ , we have

$\frac{\Phi_2 - \Phi_1}{dV} = \frac{D_y(x, y+dy, z) - D_y(x, y, z)}{dy}$

In the limit as $dx, dy, dz\rightarrow 0$ , this becomes the partial derivative in y. If we use the same procedure for the other four sides, we arrive at the definition for divergence in rectangular coordinates. (See Divergence for a full set of definitions in other coordinate systems.)

$\nabla\cdot\textbf{D} = \frac{\partial D_x}{\partial x} + \frac{\partial D_y}{\partial y} + \frac{\partial D_z}{\partial z}$ .

Going back to the surface integral definition, we see that the surface integral is just the charge (which we can call $Δ Q$ ) enclosed in the incremental (small) volume $Δ V$ . We see that

$\nabla\cdot\textbf{D} = \lim_{\Delta V\rightarrow 0}\frac{\Delta Q}{\Delta V}$ .

The expression under the limit is just an expression of the charge density $\rho\,$ at the point (x, y, z). Hence we have the first of Maxwell's Equations