66 Week 2: Continuous Charge and Gauss’s Law
We used charge per unit volume in this discussion, but we will find that charge often distributes
itself on surfaces, and we’ll often need to find the field produced bylines as well. We therefore define
all of the charge densities we might need to handle these cases as:
ρ=dq
dV⇐⇒ dq=ρ dV (36)σ=dq
dA⇐⇒ dq=ρ dA (37)λ=dq
dℓ⇐⇒ dq=ρ dℓ (38)the charge per unit volume, per unit area, and per unit length respectively. In each equation I put
the way we will need touseit – to finddq– after the defining expression.
There are thus three steps associated with solving an actual problem:a) Draw a picture, add a suitable coordinate system, identify the right differential chunk (one
you can integrate over) and draw in the vectors needed to expressdE~as given above.b) Put down an expression fordE~ (or rather, usually|dE~|) in terms of the coordinates, and
find itsvectorcomponents in terms of those same coordinates, using symmetry to eliminate
unnecessary work.c) Do the integral(s), find the fieldE~at the desired point.The first two are pretty simple, and are worth most of the credit. The last will be easy enough
if you’ve done the homework and are working hard to relearn all the calculus you need to do the
integrals required in this course, and especially at the beginning if youcan’t do the integral you
won’t be heavily penalized if you do the first two steps correctly. It’sstill something you need to
work on to get the most possible credit.
Let’s try some examples.