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The Point Source

For example, consider a hypothetical ``spherically symmetric'' sprinkler head (perhaps meant to uniformly irrigate the inside surface of a hollow spherical space colony): located at the centre of the sphere, it ``emits'' (squirts out)  dQ/dt  gallons per second of water in all directions equally, which is what we mean by ``spherically symmetric'' or ``isotropic.''^18.1 Here Q is the ``amount of stuff'' - in this case measured in gallons. Obviously (beware of that word, but it's OK here), since water is conserved the total flow of water is conserved: once a ``steady-state'' (equilibrated) flow has been established, the rate at which water is deposited on the walls of the sphere is the same as the rate at which water is emitted from the sprinkler head at the centre. That is, if we add up (integrate) the ``flux''  $\Vec{J}$  of water per second per square meter of surface area at the sphere wall over the whole spherical surface, we must get  dQ/dt.  Mathematically, this is written

$$\oSurfIntS \Vec{J} \cdot d\Vec{A} = {dQ \over dt}$$ (18.1)

where the   $\osurfintS$  stands for an integral (sum of elements) over a closed surface  ${\cal S}$.  [This part is crucial, inasmuch as an open surface (like a hemisphere) does not account for all the flux and cannot be used with GAUSS' LAW]. Now, we must pay a little attention to the vector notation: the ``flux'' $\Vec{J}$ always has a direction, like the flux (current) of water flowing in a river or in this case the flux of water droplets passing through space.

  

Figure: An isotropic source.

Each droplet has a (vector) velocity, and the velocity and the density of droplets combine to form the ``flux'' as described above. Not so trivial is the idea of a vector area element  $d\Vec{A}$,  but the sense of this is clear if we think of what happens to the scalar flux (in gallons/sec) through a hoop of wire of area  $d\Vec{A}$  when we place it in a river: if the direction of the flow of the river is perpendicular (``normal'') to the plane of the hoop, we get the maximum possible flux, namely the vector flux magnitude (the flow rate of the river) times the area of the hoop; if we reorient the hoop so that its area intercepts no flow (i.e. if the direction  $\hat{n}$  ``normal'' to the plane of the hoop is perpendicular to the direction of flow of the river) then we get zero flux through the hoop. In general, the scalar rate of flow (here measured in gallons/sec) through a ``surface element'' $d\Vec{A}$ whose ``normal'' direction $\hat{n}$ is given by $(\Vec{J} \cdot \hat{n})dA$ or just $\Vec{J} \cdot d\Vec{A}$ where we have now defined the vector surface element $d\Vec{A} \equiv \hat{n} dA$. This is pictured in Fig. 18.1 above.

Returning now to our sprinkler-head example, we have a Law [Eq. (1)] which is a mathematical (and therefore quantitative) statement of the colloquial form, which in principle allows us to calculate something. However, it is still of only academic interest in general. Why? Because the integral described in Eq. (1) is so general that it may well be hopelessly difficult to solve, unless (!) there is something about the symmetry of the particular case under consideration that makes it easy, or even ``trivial.'' Fortunately (though hardly by accident) in this case there is - namely, the isotropic nature of the sprinkler head's emission, plus the spherically symmetric (in fact, spherical) shape of the surface designated by ``${\cal S}$'' in Eq. (1). These two features ensure that

1.

the magnitude $J = \vert\Vec{J}\vert$ of the flux is the same everywhere on the surface ${\cal S}$; and

2.

the direction of $\Vec{J}$ is normal to the surface everywhere it hits on ${\cal S}$.

In this case,   $\Vec{J} \cdot d\Vec{A} = JdA$  and  J  is now a constant which can be taken outside the integral sign, leaving

$$J \oSurfIntS dA = \dot{Q}$$

where $\dot{Q}$ is just a compact notation for dQ/dt. But   $\osurfintS dA$  is just the area of the sphere,  $4 \pi r^2$,  where  r  is the radius of the sphere, so (1) becomes

$$4 \pi r^2 J = \dot{Q}$$

or

$$J(r) = {\dot{Q} \over 4 \pi r^2 }$$ (18.2)

which states the general conclusion for any spherically symmetric emission of a conserved quantity, namely

The flux from an isotropic source points away from the centre and falls off proportional to the inverse square of the distance from the source.

This holds in an amazing variety of situations. For instance, consider the ``electric field lines'' from a spherically symmetric electric charge distribution as measured at some point a distance  r  away from the centre. We visualize these electric field ``lines'' as streams of some mysterious ``stuff'' being ``squirted out'' by positive charges (or ``sucked in'' by negative charges). The idea of an electric field line is of course a pure construct; no one has ever seen or ever will see a ``line'' of the electric field $\Vec{E}$, but if we think of the strength of $\Vec{E}$ as the ``number of field lines per unit area perpendicular to $\Vec{E}$'' and treat these ``lines of force'' as if they were conserved in the same way as streams of water, we get a useful graphical picture as well as a model which, when translated into mathematics, gives correct answers. As suspicious as this may sound, it is really all one can ask of a physical model of something we cannot see. This is the sense of all sketches showing electric field lines. For every little bit (``element'') of charge  dq  on one side of the symmetric distribution there is an equal charge element exactly opposite (relative to the radius vector joining the centre to the point at which we are evaluating $\Vec{E}$); the ``transverse'' contributions of such charge elements to $\Vec{E}$ all cancel out, and so the only possible direction for $\Vec{E}$ to point is along the radius vector - i.e. as described above. An even simpler argument is that there is no way to pick a preferred direction (other than the radial direction) if the charge distribution truly has spherical symmetry. This ``symmetry argument'' is implied in Fig. 18.1.

Now we must change our notation slightly from the general description of Eqs. (1) and (2) to the specific example of electric charge and field. Inasmuch as one's choice of a system of units in electromagnetism is rather flexible, and since each choice introduces a different set of constants of proportionality with odd units of their own, I will merely state that ``J  turns into  E,   $dQ/dt \to q$  now stands for electric charge, and there is a   $1/\epsilon_0$  in front of the   $dQ/dt \equiv q$  on the right-hand side of Eq. (1)'' to give us the electrostatics version of (1):

$$\oSurfIntS \Vec{E} \cdot d\Vec{A} = {q \over \epsilon_0} %$$ (18.3)

which, when applied to the isotropic charge distribution, gives the result

$$E(r) = {q \over 4 \pi \epsilon_0} \cdot {1 \over r^2} %$$ (18.4)

The implication of Eq. (3) is then that, since the spherical shell contains the same amount of charge for all radii  r > R,  where  R  is the physical radius of the charge distribution itself, it cannot matter how the charge is distributed (as long as it is spherically symmetric); to the distant observer, the  $\Vec{E}$  field it produces will always look just like the  $\Vec{E}$  field due to a point charge  q  at the centre; i.e. Eq. (4).