How to Convert Maxwell's Equations into Integral Form

[Anne Leite]

Homework Statement

I'd like to know how to convert Maxwell's Equations from Differencial form to Integral form.

Homework Equations

Gauss' Law

Gauss' Law for Magnetism

Faraday's Law

The Ampere-Maxwell Law

The Attempt at a Solution

Convert using properties of vector analysis (as Divergence and Curl). May I convert them, using the opposite process that I would use to convert the integral form to differential form?

 

[Charles Link]

There are a couple of ways to get integral forms of Maxwell's equations. Gauss' law can be applied to the equation $\nabla \cdot E=\rho/\epsilon_o$ to give $\int \nabla \cdot E \, d^3x= \int E \cdot \, dA=Q/\epsilon_o$ where $Q$ is the total charge enclosed, and the flux of the electric field is over the surface of the enclosed volume. An alternative integral form can be written for this equation: $E(x)=\int \frac{\rho(x')(x-x')}{4 \pi \epsilon_o |x-x'|^3} \, d^3x'$ . This second form is essentially Coulomb's law (inverse square law) for the electric charge distribution $\rho(x)$. $\\$ Gauss' law can also be employed on the equation $\nabla \cdot B=0$, but usually this one is just used in a qualitative form where the lines of flux for the magnetic field $B$ are said to be continuous. In integral form, it reads $\int B \cdot \, dA=0$. $\\$ $\\$ For the curl equations, Stokes theorem ($\int \nabla \times E \cdot \, dA=\oint E \cdot \, ds$), is usually employed, e.g. Faradays law becomes $\varepsilon= \oint E \cdot \, ds=-\int (\frac{\partial{B}}{\partial{t}}) \cdot \, dA$. $\\$ For the curl B, in the steady state, again Stokes law is often employed to give $\oint B \cdot ds=\mu_o I$. An alternative integral form does exist for this one also in the steady state which is the Biot-Savart Law: $B(x)=\int \frac{\mu_o J(x') \times (x-x')}{4 \pi |x-x'|^3} \, d^3x'$. For the non-steady state, I believe the solutions of Maxwell's equations are found by the Lienard-Wiechert method, but that is likely to be beyond the scope of what you are presently doing. $\\$ It is worth noting that the integral forms of these, which is sometimes treated in courses in vector calculus, are a little more complicated than simply going from differentiation to integration.