Physics Friday 39

Let us consider two point charges of charge –q and q (q>0) separated by a distance d. Let d be the vector from the negative to the positive charge. Choosing the point halfway between the charges as the origin of our coordinate system, then the electric potential is given by:

and the electric field is

First, let us examine the field far from the charges (rd). To simplify things, let θ denote the angle between d and r. Then the law of cosines tells us that
. Thus, for rd, we see


So we have
Using , we see that for distant points, the field is approximated by
Defining , and using the unit vector in the direction of r , we have
, or, in terms of θ,
, so we see the magnitude of the distant field goes as the inverse cube of distance. The vector p is the dipole moment, and if one lets d go to zero while holding p constant, the field approaches the one above at all points, and one has a point dipole.

Now, let us instead consider the potential and field near the point between the charges (rd). In this case, , and


and so
, where is the unit vector in the direction of d. Note that this approximation is a constant vector. Thus, if we have while holding constant, our electric field appoaches a uniform electric field.


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One Response to “Physics Friday 39”

  1. Physics Friday 40 « Twisted One 151’s Weblog Says:

    […] +Q placed on the z axis at z=-a and a charge of -Q placed at z=+a, with a≫R. Then, as noted here, the field near the origin is , and if we take a,Q→∞ with held constant, the result […]

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