Skip to main content

Section 4.6 Series Expansions for Two Point Charges

To explore the behavior of the electrostatic potential due to two charges, you can expand the potential in regions of space that have convenient symmetry. You should complete at least a few cases from the following activity. (The mathematical manipulations in all of the cases are similar, but different cases will teach you different things about the physics.)

Activity 4.6.1.

Start with the formula for the electrostatic potential \(V(\rr)=V(x,y,z)\) everywhere in space due to two point charges that that you found in Activity 4.5.1.

Then find the electrostatic potential everywhere in space for the following cases:

  1. Two charges \(+Q\) and \(+Q\) are placed on a line at \(z=D\) and \(z=-D\) respectively. What is a fourth order approximation to the electrostatic potential \(V(0,0,z)\) for \(|z| \ll D\text{?}\) For what values of \(z\) does your series converge? For what values of \(z\) is your approximation a good one? Which direction would a test charge move under the influence of this electrostatic potential?

  2. As above, but for \(|z| \gg D\text{.}\)

  3. As above, but for \(V(x,0,0)\) and \(|x| \ll D\text{.}\)

  4. As above, but for \(|x| \gg D\text{.}\)

Finally, repeat all four cases above with two charges \(+Q\) and \(-Q\) placed on a line at \(z=+D\) and \(z=-D\) respectively.

Hint.

Your final answers should explicitly involve \(x,y,z\) and \(D\text{.}\)

All of the cases can be approximated by exploiting one of the common power series that we suggest you memorize; see [2]. There is no need to compute the coefficients using differentiation. In fact, if you try differentiation it will get rapidly messy!

It's crucial to understand the approximation process: Make sure you are expanding in terms of something small! This requires rewriting each term in the potential, typically to make it look like

\begin{gather*} (1 + \hbox{something small})^{\hbox{ power}} . \end{gather*}

In the series, you are adding terms of different powers. Therefore, the something small must be dimensionless, otherwise the power series is dimensionally nonsense and therefore wrong.

You may be tempted to start by adding up the potentials from the two charges, and putting them over a common denominator. In most cases, it turns out that the resulting expression is somewhat more difficult to expand in a power series than the one you started with. There is not way of knowing this in advance, you just have to try.

In some cases, the contributions from the two terms cancel. How does this cancellation relate to the symmetries of the problem?

After expanding these potentials near the origin or far away, look at the leading term or two and ask yourself if these terms are what you expect. Think especially about the symmetry or anti-symmetry of the charge distribution and the symmetry or anti-symmetry of your answer. It is very easy to ignore the sign of variables and inadvertently to drop absolute value signs. Also think about whether the leading terms give you the fall-off behavior you expect.

The power series that you use from [2] are each valid for some range of values of their dimensionless parameters. Make sure you can use this information to find where your series approximation for the potential is valid.

Answer.

For two positive charges \(+Q\) positioned at \(z=\pm D\text{:}\)

\begin{align*} V(0,0,z) \amp = {Q\over 4\pi\epsilon_0}\, {2\over D} \left( 1 + {z^2\over D^2} + {z^4\over D^4} + ...\right) \amp\text{ for} |z|\lt D\\ V(0,0,z) \amp = {Q\over 4\pi\epsilon_0}\, {2\over |z|} \left( 1 + {D^2\over z^2} + {D^4\over z^4} + ...\right) \amp\text{ for} |z|>D\\ V(x,0,0) \amp = {Q\over 4\pi\epsilon_0} {2\over D} {\left(1 - {1\over 2}{{x^2}\over {D^2}} + {3\over 8}{{x^4}\over {D^4}} + ...\right)} \amp\text{ for} |x|\lt D\\ V(x,0,0) \amp = {Q\over 4\pi\epsilon_0} {2\over |x|} {\left(1 - {1\over 2}{{D^2}\over {x^2}} + {3\over 8}{{D^4}\over {x^4}} + ...\right)} \amp\text{ for} |x|>D \end{align*}

For a positive charge \(+Q\) positioned at \(z=+D\) and a negative charge \(-Q\) positioned at \(z=-D\text{:}\)

\begin{align*} V(0,0,z) \amp = {Q\over 4\pi\epsilon_0}\, {2\over D} \left({z\over D} + {z^3\over D^3} + {z^5\over D^5} + ...\right) \amp\text{ for} |z|\lt D\\ V(0,0,z) \amp = {Q\over 4\pi\epsilon_0}\, {2\over |z|} \left({D\over z} + {D^3\over z^3} + {D^5\over z^5} + ...\right) \amp\text{ for} |z|>D\\ V(x,y,0) \amp = 0\amp\text{ for} (x,y,0) \end{align*}