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The Line Source

A sphere, as we have seen, can be collapsed to a point without affecting the external field; and a point is essentially a ``zero-dimensional object'' - it has no properties that can help us to define a unique direction in space. The next higher-dimensional object would be one-dimensional, namely a line. What can we do with this?

In the spirit of the normal physics curriculum, we will now stick to the example of electrostatics, remembering that all the same arguments can be used on gravity or indeed on other situations not involving ``force fields'' at all. (Consider the sprinkler, or a source of ``rays'' of light.) Suppose that we have an ``infinite line of charge,'' i.e. a straight wire with a charge  $\lambda$  per unit length. This is pictured in Fig. 18.2.

The same sort of symmetry arguments used in Fig. 18.1 tell us that for every element of charge a distance  d  above position  x  on the wire, there is an equal element of charge an equal distance  d  below position  x,  from which we can conclude that the ``transverse'' contributions to the  $\Vec{E}$  field from the opposite charge elements cancel, leaving only the components pointing directly away from the wire; i.e. perpendicular to the wire. In what are referred to as ``cylindrical coordinates,'' the perpendicular distance from the wire to our field point is called  r,  and the direction described above is the  r  direction. Thus  $\Vec{E}$  points in the  $\hat{r}$  direction. (Indeed, if it wanted to point in another direction, it would have to choose it arbitrarily, as there is no other direction that can be defined uniquely by reference to the wire's geometry!) Given the direction of  $\Vec{E}$  and the ``obvious'' (but nevertheless correct) fact that it must have the same strength in all directions (i.e. it must be independent of the ``azimuthal angle''  $\phi$  -- another descriptive term borrowed from celestial navigation), we can guess at a shape for the closed surface of Eq. (3) which will give us  $\Vec{E}$  either parallel to the surface (no contribution to the outgoing flux) or normal to the surface and constant, which will let us take  E  outside the integral and just determine the total area perpendicular to  $\Vec{E}$:  we choose a cylindrical shaped ``pillbox'' centred on the wire. No flux escapes from the ``end caps'' because  $\Vec{E}$  is parallel to the surface;  $\Vec{E}$  is constant in magnitude over the curved outside surface and everywhere perpendicular (normal) to it. Thus

$$\oSurfIntS \Vec{E} \cdot d\Vec{A} = E \oSurfIntS dA = (E)(2 \pi r L)$$

where  $2 \pi r L$  is the curved surface area of a cylinder of radius  r  and height  L.

  

Figure: An infinite, uniform line of charge.

The same surface, clipping off a length  L  of wire, encloses a net charge   $q = \lambda L$.  Plugged into (3), this gives

$$2 \pi r L E = {\lambda L \over \epsilon_0} %$$

or

$$E(r) = {\lambda \over 2 \pi \epsilon_0} \cdot {1 \over r} %$$ (18.8)

which states the general conclusion for any cylindrically symmetric charge distribution, namely that

The electric field from a cylindrically symmetric charge distribution points away from the central line and falls off proportional to the inverse of the distance from the centre.

This also holds in an amazing variety of situations. Applications are left to the interested student.