On Problem 2.2 in Griffiths' Introduction to Electrodynamics

Question

In part (a) we were asked to find the electric field at a distance $z$ above the midpoint between two equal charges of magnitude $q$ that are a distance $d$ apart. I obtained the correct answer:

$$\frac{1}{4\pi \epsilon_0}\frac{2qz}{\left[z^2+{\left(\frac{d}{2}\right)^2}\right]^{\frac{3}{2}}} \hat{z}.$$

We were asked to verify whether the result is consistent with what one would expect for $z \gg d$. The answer is:

$$\frac{1}{4\pi \epsilon_0}\frac{2q}{z^2},$$

and the explanation given is that at a very high $z$ the system appears to be a single point charge of magnitude $2q$. Fair enough.

My question is about part (b), where we are asked to repeat the problem for a system of charges $q$ and $-q$. The answer I derived is correct. Although I do understand that $z \gg d$ is the case of a short dipole, I wonder why the argument used in part (a) cannot be extended to this problem and why we can't conclude that the field at $z \gg d$ is zero. Any help would be appreciated.

Answer 1

When you get to chapter 3 of Griffiths you'll find out that at large distances, the electric field can be written in a multipole expansion, which is effectively a power series in $1/r$: $$ \vec{E} = \frac{1}{r^2} \left( \text{monopole piece} \right) + \frac{1}{r^3} \left( \text{dipole piece} \right) + \frac{1}{r^4} \left( \text{quadrupole piece} \right) + \cdots $$ The "monopole moment" is just another term for the total charge of the configuration, and you might have already heard of the "dipole moment" of a charge configuration; it's a vector that tells you how much positive and negative charges are separated in a charge configuration. In part (b) of this problem case, it has magnitude $qd$ and points from the negative to the positive charge.¹

At very large distances, the dominant behavior of the electric field will be determined by the first term in this series that isn't exactly zero. In part (a) of the problem, the system has a net charge (a monopole moment) of $2q$, and so we expect that the electric field will be proportional to $1/r^2$. But in part (b), the net charge is zero, and so we say that "the electric field vanishes to first order" in this series. To find out what the dominant behavior of the electric field is when $r \gg z$, we have to go to the next order in the series, the dipole term. Since we know that the dipole moment of this configuration is non-zero, we would expect the field to be proportional to $qd/r^3$ in this limit. It is left as an exercise for the reader to confirm that this is true.

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¹ Unless you're a chemist.