Let's compute what this electrostatic potential V is, in the case where you are
standing at a distance d, away from the positive charge, and where d is much
greater than r, the separation distance between the two charges in your dipole.
Now I'm going to assume that we are orthogonal to the dipole, that is we're
at distance d from the positive charge but perpendicular to the orientation of
the dipole. That means that the distance to the
negative charge is, by Pythagoras's theorem, the square root of d squared
plus r squared. So that in computing V, we get kq over d,
the positive charge distance, minus kq over square root of d squared plus r
squared. We could stop there, but let's simplify
this as much as possible. If we factor out kq over d, then we get a
one from the left term, from the right term we get 1 over the square root of
quantity 1 plus r over d quantity squared.
Well, at this point we can try to expand that out using the binomial series.
Recall the binomial series is for something of the form 1 plus x to the
alpha. In this case, alpha is negative one half.
We're looking at one over the square root of something, and x is r over d quantity
squared. Because the first two terms in the
binomial series are 1 plus alpha x, we get the first two terms in the series
applied here to be 1 minus one half times r over d quantity squared.
Simplifying, we see that the ones cancel and we're left with kq over d times one
half, are over d quantity squared. Everything else is of higher order term
in r squared over d squared. Let's remember that one half r over d
quantity squared. In fact, we'll set it over here, right
next to the location. And consider what happens when we look at
distance d away from that positive charge, but in a different direction.
Instead of being orthogonal to the dipole, let's look parallel to it.
Well, in this case, d plus remains the same.
It is this distance d. But now d minus is a bit simpler looking.
It is d plus r, the separation distance in the dipole.
Now again, factoring out a kq over d, we get a term on the right hand side that is
of the form, one over one plus r over d. We can expand this out using, if we like,
the binomial series with alpha equal negative one, or what that really is, the
geometric series. For 1 over 1 minus x, with x being equal
to negative r over d. Writing out the first two terms of that
series, as well, we get 1 minus r over d. The ones cancel, just as before.
And we see that up to the leading order term, our electrostatic potential is kq
over d times r over d. Now, notice how different that is from
before. By using different series, we see that
the leading order terms in the electrostatic potential differ depending
on how you are oriented with respect to the dipole.
Inline potential is much stronger than orthogonal.