W9_parallel_resonance.eps

Week 5: Resistance 167

Let’s carefully examine the picture and see what we can deduce. We are interested in computing
theelectric current, defined to be thecharge per unit timethat is being carried by the conduc-
tor. Ordinarily, we’d think of this as the charge per unit time travellingin some chosen direction that
passes some point in the conducting wire under the influence of the force created by the battery (or
other source of potential difference across the wire). However,what does “passing a point” mean?
How can we manage our choice of direction? All of the charge may notbe travelling in the same
direction! The conductor may not be a simple cylinder like that pictured above but instead be some
contorted shape cast in metal with many branches! We need a better, less ambiguous definition.

Wecanunambiguously estimate how much charge passes through a givensurfacecutting across
the metal. Since a surface in three dimensions has one dimension perpendicular to the surface, we
can always assign our direction unambiguously in this perpendicular direction. The wise student
should already be saying to themselves “But that sounds a lot like ourreasoning when we talked
about the flux of the electric field a few chapters back..” and that wise student would be quite right!

But first things first. For the time being, let’s confine ourselves to the simple case of the cylinder
above with the surface in question being one that cuts across itat right angles, the surfaceA
pictured above. From the picture we can see thatall of the charge∆Qin the volume between the
plane surface bounded by the dashed circle and the plane surfaceAbounded by the circle at the
far right of the conductor passes through the cross-sectionalareaAperpendicular to the direction
of motion of the charges in a time ∆t. So how much is that?

To answer this, we need to define a few quantities. One is:

n=

# of charge carriers

unit volume

(288)

the number of (free!) charge carriersqper unit volume. We can then turn this into thefree charge
density:
ρfree=nq (289)

Using these quantities, we see that:

∆Q=nqvdA∆t=ρfreevdA∆t (290)

which we read as “the number of charge carriers per unit volume times the charge per carrier times
the volumevdA∆t”. This means (dividing out the ∆t) that the total charge per unit time that goes
throughAis:

I=

∆Q

∆t

≈

dQ

dt

=nqvdA=ρfreevdA (291)

In passing we note that the SI units of current areAmperes(or Amps for short) where

1 Ampere =1 Coulomb

1 Second

(292)

The resultI=nqvdAwill occur again and again when we pass from a microscopic descriptionof
e.g. magnetic forces on charges to macroscopic forces on current carrying wires, so keep it in mind!
It isn’t just a transient “use once” result; it is the key to understanding many things.

Current Density and Charge Conservation

Note well that in the picture above, we determine the current thatpasses “a point in the wire” by
evaluating how much charge passes through somesurfacethat contains the point! The particular
surface we chose in our simple derivation is one perpendicular to the direction of the motion of
the charge, but we cannot possibly guarantee that all conductors carrying a current will have some
simple known surface where this is true. Also, as we noted above, this picture should remind you of