W9_parallel_resonance.eps

Week 2: Continuous Charge and Gauss’s Law 65

Now we can sum over a lotfewerobjects. There aren’t as many blocks a micron in size as there
were charges, but there arestillway, way too many blocks in an object even the size of a centimeter

• 10^12 of them, in fact – too many for us to actually sum up with a calculator.Generally, however,
  ρvaries only alittlefrom block to block. Also, on a centimeter-plus scale, those micron sized blocks
  areinfinitesimal, small enough to treat as if they aredifferentialin size. We can then consider using
  calculusto do our sums. Here’s how it works:

r

r − r

r

P

i

i

Figure 9: Coarse grained average leading to an integral.

In the amoebic blob shaped object above, we’ve chopped the whole volume up into little chunks
∆V in size (highly exaggerated in the picture so you can see them). We’vetallied up the charge in
each block ∆Q, and labelled (in our minds) each block with an indexiat position~ri. We can then
compute the field using the superposition principle at the pointP(position~r) as:

E~tot(~r) =

∑

i

k∆Qi

|~r−~ri|^2

(~r̂−~ri) (32)

As noted, there are too many chunks in the blob for us to sum over. So we pretend that the
charge iscontinuously distributedaccording to:

ρ= lim∆V→ 0

∆Q

∆V =

dQ

dV (33)

and turn the summation into anintegral(remember bothσand

∫

stand for S(um), they are both
summation symbols, the latter the one we use for continuous things):

E~tot(~r) =

∑

i

k∆Qi

|~r−~ri|^2

(~r̂−~ri) =

∫

V

kρ(~r′)dV′

|~r−~r′|^2

(~r̂−~r′) (34)

where we’ve useddQ=ρdV (in the primed coordinates we use to replace the~ri’s). This is just
thefield of every little differential sized chunk that makes up the entire object, summed over all the
chunks!

This is a lot to remember, so we’ll create a little mnemonic to help you. Just as we found
the electric field last week by using the field of a single point charge in itssimplest form and then
putting it into suitable coordinates, we’ll find it this week the exact same way, but the point charge
in question will bedqand notq. That is:

E~=kq

r^2

ˆr ⇐⇒ dE~=

k dq

r^2

rˆ (35)

To use the latter, we just have to finddqfor the particular kind of distribution, and be able to do
the final integrals.