Electric flux

The flux of a vector field is a quantity that reflects how much of the field is passing through an area. The fourth equation on the AP Physics C Electricity & Magnetism equation sheet gives the electric flux: \[\Phi_E = \iint \vec{E} \cdot d\vec{A}.\qquad{(1)}\]

In eq. 1, the notation \(d\vec{A}\) means \(\hat{n} dA\), where \(\hat{n}\) is the normal to \(dA\), so that the fluxfrom fluxus, from Latin, fluere, “to flow”

quantifies how much field is flowing through the area, normal to it. The dot product with the normal means we do not count stuff passing parallel to the area, which doesn’t really flow through it. We will see similar fluxes for magnetic fields and current. Flux also occurs in other fields like heat and mass transfer, nuclear reactor theory, fluid mechanics, and more.

The \(\iint dA\) notation means we are doing a double integralYou probably are not yet doing these in your multivariate calculus class, but you will. Typical applications of double integrals are to find areas (\(\iint dA\)), moments of area (\(\iint r^2 dA\)), surface areas, cumulative probabilities under multivariate pdfs, etc. There are also triple integrals, often used to find volumes (e.g. \(\iiint dV\)).

over an area. You are used to doing a single integral, e.g. \(\int dx\). A double integral could be the integral of an integrand that itself is an integral, such as \(\int \int dx dy\), perhaps rewritten as \(\iint dA\) to be less cumbersome. Sometimes, in the interest of writing quickly, we may write \(\int dA\), recognizing it as a double integral because it is with respect to \(dA\).

Integral form of Gauss Law

Gauss’s lawDeveloped some by Joseph-Louis Lagrange in 1773; mainly by Carl Freidrich Gauss in 1835. See also https://en.wikipedia.org/wiki/Gauss\%27s_law. Gauss’s law is the first of four Maxwell’s Equations which describe electromagnetic fields; we’ll get to the other three later.

relates the electric flux out of an arbitrary closed contour to the charge enclosed (\(q_{enc}\)) within the countour: \[\begin{aligned} \oiint \vec{E}\cdot d\vec{A} &= 4\pi k q_{enc} \\ &= \dfrac{1}{\epsilon_0} q_{enc} \end{aligned}\qquad{(2)}\]

The notation \(\oiint d\vec{A}\) means the flux is taken over a closed contour. The contour can be any conveniently chosen shape so long as it has no holes and doesn’t leak. You could choose to integrate over the surface of a potato shape; more often however, the contour is a well-chosen sphere or cylinder.

In eq. 2, \(q_{enc}\) is the charge enclosed within the contour. The actual distribution of the charge within the contour does not really matter for application of eq. 2.

Gauss’s law is a bit funny in that it doesn’t work directly with \(\vec{E}\) but rather it is a statement about the double integral of \(\vec{E}\) over some closed contour. At first sight, these would seem to leave us worse off by adding all kinds of confusing advanced math to the stuff we must think about. In practice, however, we can sneak around the advanced math by making wise choices of the closed contour we integrate over, turning the multivariate calculus into a geometry problemIn 6.003 Signals and Systems, MIT EECS Prof Stephen Senturia used to say “We don’t want to do math! We suck at math! We make stupid mistakes when we try to do math!” suggesting we should find clever ways to get around actually doing the math. Compare to Chinese warrior philosopher Sun Tzu (544-496 BCE), “The supreme art of war is to subdue the enemy without fighting.” Checkmate.

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For now, we will start using Gauss’s law without deriving it. Gauss’s law can be developed from Coulomb’s law using some multivariate calculus sleight of hand and the sifting properties of the Dirac delta function, but we will not yet go there because you have yet to get to these in your multivariate calculus class.

Differential form of Gauss’s Law (optional aside)

eq. 2 is the integral form of Gauss’s law; there is also a differential form: \[\vec{\nabla}\cdot\vec{E}=\dfrac{\rho}{\epsilon_0}.\qquad{(3)}\] The differential form is often put on t-shirts at fancy engineering schools but is not in any way testable in AP Physics C Electricity and MagnetismIt is mentioned here merely for those of you who love seeing odd math and new math symbols so much that it makes you want to go out and know what the heck these things mean… You may see this in an physics (e.g. 8.022) or multivariate calculus (e.g. 18.023) class next year.

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While the integral form is useful when symmetry and geometry in the problem allows us to pick a convenient contour, the differential form is more usable otherwise. The fancy \(\vec{\nabla}\cdot\vec{E}\) notation means the divergence of \(\vec{E}\), a multivariate calculus operation that works kind of like our flux calculation as we shrink our closed contour to be very tiny. There are several multivariate calculus theorems that become useful when relating such quantities to path, area, and volume integrals which they will go over later. The upside-down triangle\(\vec{\nabla}\) is typically spoken as “del”

\(\vec{\nabla}\) is a multivariate differential calculus vector operator given in Cartesian coordinates by \[\vec{\nabla}=\dfrac{\partial}{\partial x}\hat{i} + \dfrac{\partial}{\partial y}\hat{j} + \dfrac{\partial}{\partial z}\hat{k}.\]

Thin cylindrical shell with charge density \(\sigma\)

We choose a cylindrical contour of radius \(r\). There are two solutions.

Where \(r<R\), \(q_{enc}=0\) so by inspection \(\vec{E}=0\) for \(r<R\).

Where \(r>R\), the solution should look just like our line charge solutionIn physics, it is nice to turn a problem into one you already know how to do.

, with \(\lambda = \sigma 2\pi R\); but let’s do it out anyway.

\[\begin{aligned} \oiint\vec{E}\cdot d\vec{a} &= 4\pi k q_{enc} \\ |\vec{E}| \iint_\text{cylinder} da &= 4\pi k q_{enc} \\ |\vec{E}| 2\pi r l &= 4\pi k q_{enc} \\ |\vec{E}| 2\pi r l &= 4\pi k \sigma 2\pi R l \\ |\vec{E}| &= 4\pi k \sigma \dfrac{R}{r} \\ \vec{E}(r) &= \begin{cases} 0 & r < R \\ 4\pi k \sigma R \dfrac{1}{r} \hat{r} = \dfrac{\sigma R}{\epsilon_0} \dfrac{1}{r} \hat{r} & r \geq R \end{cases} \end{aligned}\]

Solid cylinder with charge density \(\rho\)

⊕{#fig:gausslawsolidcylinder} Gauss’s law contour for electric field for a solid cylinder with charge density \(\rho \unit{\coulomb\per\meter\cubed}\).

For a solid cylinder of radius \(R\) with charge density \(\rho\) C m⁻³, the solution for \(r>R\) should look identical to the solutions for a line charge (\(\lambda=\pi R^2 \rho\)) or a cylindrical shell (\(\sigma = \frac{R}{2}\rho\)).

\[\begin{aligned} \oiint\vec{E}\cdot d\vec{a} &= 4\pi k q_{enc} \\ |\vec{E}| \iint_\text{cylinder} da &= 4\pi k q_{enc} \\ |\vec{E}| 2\pi r l &= 4\pi k q_{enc} \\ |\vec{E}| 2\pi r l &= 4\pi k \rho \pi R^2 l \\ |\vec{E}| &= 2\pi k \rho \dfrac{R^2}{r} \\ \end{aligned}\]

For \(r<R\), inside the solid cylinder, \[\begin{aligned} \oiint\vec{E}\cdot d\vec{a} &= 4\pi k q_{enc} \\ |\vec{E}| \iint_\text{cylinder} da &= 4\pi k q_{enc} \\ |\vec{E}| 2\pi r l &= 4\pi k q_{enc} \\ |\vec{E}| 2\pi r l &= 4\pi k \rho \pi r^2 l \\ |\vec{E}| &= 2\pi k \rho r \\ \end{aligned}\]

The electric field is \[\vec{E}(r) = \begin{cases} 2\pi k \rho r \hat{r} & r < R \\ 2\pi k \rho R^2 \dfrac{1}{r} \hat{r} & r \geq R \end{cases}\]

Thin spherical shell with charge density \(\sigma\)

⊕{#fig:gausslawsphericalshell} Gauss’s law contour for electric field for a thin spherical shell with charge density \(\sigma \unit{\coulomb\per\meter\squared}\).

Choose a Gauss’s law contour that is a concentric sphere of radius \(r\). Where \(r>R\) the solution will look like a point charge with \(Q=\sigma 4\pi R^2\). Inside the spherical shell, \(q_{enc}=0\) so by inspection \(\vec{E}=0\) for \(r<R\).

\[\vec{E}(r) = \begin{cases} 0 & r < R \\ \dfrac{4 \pi R^2 \sigma k}{r^2}\hat{r} & r \geq R \end{cases}\]

Solid sphere \(\rho\)

⊕{#fig:gausslawsphere} Gauss’s law contour for electric field for a solid sphere with charge density \(\rho \unit{\coulomb\per\meter\cubed}\).

It is left as an exercise for the student to show that

\[\vec{E}(r) = \begin{cases} \dfrac{4 \pi \rho k}{3} r \hat{r} & r < R \\ \dfrac{4 \pi R^3 \rho k}{3 r^2}\hat{r} & r \geq R \end{cases}\]