Simple Nature - Light and Matter

f/The flux through a tiny cube
due to a point charge.

by considering its point charges individually, and the additivity-by-

region property tells us that if we have a single point charge outside

a big region, we can break the region down into tiny cubes. If we

can prove that the flux through such a tiny cube really does cancel

exactly, then the same must be true for any region, which we could

build out of such cubes, and any charge distribution, which we can

build out of point charges.

For simplicity, we will carry out this calculation only in the spe-

cial case shown in figure f, where the charge lies along one axis of

the cube. Let the sides of the cube have length 2b, so that the area

of each side is (2b)^2 = 4b^2. The cube extends a distancebabove,

below, in front of, and behind the horizontalxaxis. There is a dis-

tanced−bfrom the charge to the left side, andd+bto the right

side.

There will be one negative flux, through the left side, and five

positive ones. Of these positive ones, the one through the right side

is very nearly the same in magnitude as the negative flux through

the left side, but just a little less because the field is weaker on

the right, due to the greater distance from the charge. The fluxes

through the other four sides are very small, since the field is nearly

perpendicular to their area vectors, and the dot productEj·Ajis

zero if the two vectors are perpendicular. In the limit wherebis

very small, we can approximate the flux by evaluating the field at

the center of each of the cube’s six sides, giving

Φ = Φleft+ 4Φside+ Φright

=|Eleft||Aleft|cos 180◦+ 4|Eside||Aside|cosθside

+|Eright||Aright|cos 0◦,

and a little trig gives cosθside≈b/d, so

Φ =−|Eleft||Aleft|+ 4|Eside||Aside|

b

d

+|Eright||Aright|

=

(

4 b^2

)

(

−|Eleft|+ 4|Eside|

b

d

+|Eright|

)

=

(

4 b^2

)

(

−

kq

(d−b)^2

+ 4

kq

d^2

b

d

+

kq

(d+b)^2

)

=

(

4 kqb^2

d^2

)(

−

1

(1−b/d)^2

+

4 b

d

+

1

(1 +b/d)^2

)

.

Using the approximation (1 +)−^2 ≈ 1 − 2 for small, this becomes

Φ =

(

4 kqb^2

d^2

)(

− 1 −

2 b

d

+

4 b

d

+ 1−

2 b

d

)

= 0.

Thus in the limit of a very small cube, b d, we have proved

that the flux due to this exterior charge is zero. The proof can be

644 Chapter 10 Fields