Simple Nature - Light and Matter

|E|=

kqtotal

r^2

,

whereris the radius of Flatcat. The flux is then

Φ =

∑

Ej·Aj,

and since theEjandAjvectors are parallel, the dot product equals

|Ej||Aj|, so

Φ =

∑kqtotal

r^2

|Aj|.

But the field strength is always the same, so we can take it outside

the sum, giving

Φ =

kqtotal

r^2

∑

|Aj|

=

kqtotal

r^2

Atotal

=

kqtotal

r^2

4 πr^2

= 4πkqtotal.

Not only have all the factors ofrcanceled out, but the result is

the same as for a disk!

Everyone is pleasantly surprised by this apparent mathematical

coincidence, but is it anything more than that? For instance, what

if the charge wasn’t concentrated at the center, but instead was

evenly distributed throughout Flatcat’s interior volume? Newton,

however, is familiar with a result called the shell theorem (page 102),

which states that the field of a uniformly charged sphere is the same

as if all the charge had been concentrated at its center.^10 We now

have three different assumptions about the shape of Flatcat and the

arrangement of the charges inside it, and all three lead to exactly the

samemathematical result, Φ = 4πkqtotal. This is starting to look

like more than a coincidence. In fact, there is a general mathematical

theorem, called Gauss’ theorem, which states the following:

For any region of space, the flux through the surface equals

4 πkqin, whereqinis the total charge in that region.

Don’t memorize the factor of 4πin front — you can rederive it

any time you need to, by considering a spherical surface centered on

a point charge.

(^10) Newton’s human namesake actually proved this for gravity, not electricity,
but they’re both 1/r^2 forces, so the proof works equally well in both cases.
642 Chapter 10 Fields