# Why can we use Gauss' law to compute electric field?

For simplicity I'm considering only the sphere case.

In the Gauss' Law formulation we have some field $E$ introduced by charges $Q$ inside some sphere, then we compute flux and integrate, and we get result $Q/e_0$. Right. But this formulation doesn't take into account any possible outer charges, because $E$ used in this law comes only from $Q$. I suppose that any outer field effects will cancel out, because flux coming from it must go through surfaces with opposite normals. But we didn't make any proof of that, as far as I have seen in Gauss' Law proofs.

So, when we measure electric field inside the sphere with uniformly distributed $Q$, we make some assumptions on uniformity of $E$ inside this sphere and we consider hollow Gauss' sphere with radius smaller than radius of the outer sphere. And then we use $E$ to compute total flux going through inner sphere and conclude that it must be zero. Right.

But why can we use this $E$? We don't know that there aren't any effects from outer charges fields. So why can we use Gauss' Law?

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## 3 Answers

The electric field used in Gauss's law is not just the electric field from the interior charge distribution. It may contain contributions from exterior sources. Usually, we do assume that there are no exterior sources, which is what allows us to impose symmetry considerations.

Why are we able to do this? Because Maxwell's equations are linear, and as such, the sum of two solutions to Maxwell's equations is itself a solution. Hence, you can consider the exterior and interior charge densities separately. You can calculate the fields separately and then add them up at the end.

Here's a mathematical way of looking at it. You have Gauss's law in integral form:

$$\int_V \frac{\rho}{\epsilon_0} \, dV = \oint_S E \cdot dS$$

Linearity of Maxwell's equations means that you can split up $\rho = \rho_\text{int} + \rho_{\text{ext}}$ and the same for $E$. The two parts must separately obey this equation. So that means we have

\begin{align*} \int_V \frac{\rho_\text{int}}{\epsilon_0} \, dV &= \oint_S E_\text{int} \cdot dS \\ \int_V \frac{\rho_\text{ext}}{\epsilon_0} \, dV &= \oint_S E_\text{ext} \cdot dS \end{align*}

But $\rho_{\text{ext}} = 0$ everywhere in $V$, so the second RHS integral must be zero.

When we do Gauss's law problems, we implicitly look at the $E_\text{int}$, the field generated by charges internal to the volume. Any external field and charges can be separated out in this way, so that the symmetry of the field generated by internal charges is still manifest, allowing us to compute the field magnitude thanks to that symmetry. It's true that, strictly speaking, we can't draw any conclusions about the total $E$-field without some information about exterior charges or fields, but by linearity, we know that our field from the internal charges will never be affected by such information.

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First of all we use Gauss's law because it is a law of nature!

Second: Gauss's law (together with $\nabla\times\vec{E}=0$, which is only strictly valid in electrostatics) is equivalent to Coulomb's law for the electric field of a distribution of charges. You can prove one from the other.

Third: You a right about external fields cancelling. Gauss's law essentially says that electric field lines only end on charges. So if you have a volume that doesn't contain a charge, any field line which enters will also leave and the total flux through the closed surface is zero.

Fourth: In the sphere problem is the sphere a conductor? If so it screens anything happening outside the sphere. If the sphere is not a conductor then the field inside will depend on what is happening outside, but the total flux through the sphere will only depend on what's inside.

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To the person downvoting all my old answers lately: if you care at all about the quality of this site then leave a reason for your downvote! I love constructive criticism. If I've said something wrong, I want to be corrected. But anonymous downvoting achieves nothing. – Michael Brown Aug 13 '13 at 1:58

The electric field under the integral in Gauss's law is the electric field of all the charged particles in the Universe. The flux on the other side of Gauss's law is the flux of only the charged particles inside your Gaussian surface. As you correctly state, the flux from the charged particles outside the Gaussian surface will indeed be zero because the fields of these particles pierces the Gaussian surface in two opposite directions. This can be easily proven using solid angle. Matter & Interactions by Chabay and Sherwood has a nice discussion of this.

The electric field inside the Gaussian surface need not be uniform. A single particle's electric field isn't uniform. Instead, the field must display a certain symmetry in order for Gauss's law to be useful. Gauss's law is always true, in that the integral of the normal component of E over any closed surface is the quotient of the enclosed charge and vacuum permittivity but this isn't always useful for working backwards to deduce E. There must be symmetry present, and it's almost always radial (spherical) symmetry, plane symmetry, or cylindrical symmetry.

To summarize, the integral in Gauss's law accounts for every charged particle in the Universe, but the other side of the equation limits the net flux only to charged particles inside a closed surface, the same surface over which the integral is evaluated. In the presence of appropriate symmetry, the integral reduces to a simple expression from which we can deduce E.,

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