Common Mistakes

Assignment #1

Tip for dealing with integrals over the volume, i.e.:

\[Q_{encl} = \int_0^a \rho(r) d\tau\]

Short version of my tip:

If the the function you’re integrating over (e.g. charge density \(\rho(r)\)) only depends on \(r\) then don’t integrate over both \(\phi\) and \(\theta\), just say

\[Q_{encl} = \int_0^a \rho(r) 4\pi r^2 dr\]

Long version of my tip:

Lots of you said (on #6 on the HW):

\[Q_{encl} = \int_0^a\int_0^\pi \int_0^{2\pi} \rho(r) r^2 \sin\theta d\theta d\phi dr\]

which is absolutely correct, but you totally don’t need to go through that trouble.

For spherically symmetry functions

\[d\tau = 4\pi r^2 dr ~~~~~{\rm For~spherically~symmetric~functions.}\]

\(4\pi r^2 dr\) is a perfect, differentially small volume element, as long as whatever you’re integrating over only depends upon \(r\) and not on \(\theta\) and \(\phi\). In other words, as long as your problem has spherical symmetry.

Another version I saw was

\[Q_{encl} = \int_0^a \rho(r) dr ~~~~~{\rm this~is~wrong}\]

That version is actually incorrect. \(\rho(r)\) is a charge per volume, so in order to get charge you need to integrate over the volume. You can get a clue about whether you’re integrating over something reasonable by looking at its units. \(dr\) only has units of length, so it can’t be a volume. \(4 \pi r^2 dr\) has units of length cubed, so it’s a volume.

Whenever I’m “setting up integrals” which I think is physics speak for “making stuff up that looks like an integral and will hopefully give me the answer I want” I look carefully at the units of the thing. The units have to work out in the integral for the units to work out in the problem. \(dr\) has units of length.