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Electric field far away from a bunch of charges

2 votes

I am given a bunch of charges and asked for the electric field at a field point far away (see image below).

What I would do is the following:

What I have done is applying Gauss' Law. However, in the lecture, another approach was used.

Firstly, the charges were treated per pairs , as dipoles. And as there are 4 +ive and 4 -ive, they cancel each other out and there is no first order contribution (i.e. $\frac{1}{r}$) .

Secondly, it was argued that this a quadrupole (using the same logic that above). Thus, the final answer would qualitatively be:

$$E \propto \frac{1}{r^3}$$

Thus using two different methods one gets different answers; using method 1 one gets $E \propto \frac{1}{r^2}$ and using method 2 one gets $E \propto \frac{1}{r^3}$

Who is wrong and why?


asked Feb 24, 2019 in Physics Problems by Jorge Daniel (716 points)
edited Mar 26, 2019 by sammy gerbil

1 Answer

2 votes
Best answer

You have not defined the problem well. "A bunch of charges" is a vague description. Both answers could be correct, or neither, depending on the arrangement of charges.

The electric field outside of any general system of charges can be modelled as a multi-pole expansion of the form $$E=k(\frac{q_0}{r^2}+\frac{q_1}{r^3}+\frac{q_2}{r^4}+...)$$ The coefficients $q_k$ depend on the polar and azimuthal angles $\theta, \phi$ and are related to the spherical harmonic functions. The 1st term represents a monopole with the resultant charge on the bunch. The 2nd and 3rd terms represent a dipole and quadrupole respectively.

If the "bunch of charges" has an overall non-zero charge $q_0$ then far from the bunch the 1st term will dominate. It will resemble a point charge with an electric field which varies approximately as $1/r^2$.

However, if the bunch has an overall zero charge $q_0=0$ but the centres of +ve and -ve charge do not coincide then the 2nd term will dominate. The electric field far from the bunch is will resemble that of a dipole, which varies as $1/r^3$.

If the centres of +ve and -ve charge do coincide, then the 2nd term will also vanish ($q_1=0$) - for example, 4 charges at the corners of a square, alternating. This arrangement could have a non-zero 3rd term (a quadrupole, $q_2 \ne 0$), for which the electric field is proportional to $1/r^4$.

answered Feb 25, 2019 by sammy gerbil (28,896 points)
edited Mar 9, 2019 by sammy gerbil