# What is the location of point charge with respect to the original charged body? [duplicate]

we say if distance between two charged bodies is large as compared to its size then we take them as point charges and assume all its charge content to be concentrated at one point in space. What is the location of that one point with respect to the original body? Is it always at the geometric centre of the body?

• Hi. A thought: If the distance between the two bodies is large compared to their size, does it make a difference if the point were we take the charge of one of the two is at point P or P+a, where a is a number very small compared to the distance between the two charges? I am phrasing this as a question. – Constantine Black Feb 20 '16 at 16:48
• See also physics.stackexchange.com/q/8221 for calculation with higher precision. – Constantine Black Feb 20 '16 at 16:50
• Also for details in the multipole expansion see en.wikipedia.org/wiki/… – Constantine Black Feb 20 '16 at 16:53
• Yeah i guess it does not make any difference but does the point depend on the charge distribution, size, shape etc of the object like centre of mass if yes in what way?? – Matt Feb 20 '16 at 17:10
• It depends on the charge distribution. Just wrote an answer if you are interested still. – Constantine Black Feb 20 '16 at 17:48

I will try to present how we can talk about the radiation of a distribution of charges around the (0,0,0) of the axes.

First, notation: r' is the distance from 0 of the axes of an infinitesimal volume containing charge $dq = ρ() dV$ where ρ is the function of charge density. Also r is the distance from the beginning of the axes of our point of observation- we measure the potential V() there. Also we shall denote by ι the distance between the infinitesimal volume of charge and the point of observation. We note that $i= \sqrt{r^2 +r' ^2 -2 \bar r \cdot \bar r'}$

So, we work with a retarded potential $V( \bar r, t)= {1 \over {4 π ε_0 } } \int {{ρ( \bar r' , t -ι/c)} \over {ι}} dV$

We shall make the following approaches:

1) r'<< r

This enables us to write: $ι= r(1 - {{ \bar r \cdot \bar r' } \over {r^2}})$ and so we have for the charge density: $ρ()=ρ( \bar r' , t-r/c + { {\bar r \cdot \bar r'} \over {c}} )$ By taking a Taylor expansion for the above function around a value $t_0=t - r/c$, the value of the retarded time, we conclude to:

$$ρ= ρ(\bar r' ,t_0) + \dot ρ( \bar r', t_0)(\hat r \cdot r' /c) + ...$$ where the dot above means derivation over time. The following terms can be neglected if we make a second approach:

2)$r'<< {{c} \over |{\ddot ρ / \dot ρ}^(1, 1/2, 1/3,...)| }$

Thus we may conclude to:

$$V(\bar r , t) = (1/4πε_0)[\int {ρ(\bar r' ,t )dV'} + (\hat r /r) \int {\bar r' ρ( \bar r' ,t_0 )dV'} + \hat r/c {{d} \over {dt}} \int {\bar r' ρ(\bar r', t_0)dV'} ]$$. That is

$$V(\bar r ,t) = (1/4πε_0)[(Q/r) + (\hat r \cdot \bar p(t_0) /r^2) + ({\hat r} \cdot {\dot {\bar p(t_0)}} /rc)]$$. Note that the last term is zero.

The two first terms are the monopole(the term that you are speaking in your question as the point where we find the 'charge') and dipole terms of the multipole expansion. For many other terms we get higher contributions.

Note that p is the electric dipole moment.

Hope this helps.