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Section 9.6 Potential Due to a Uniformly Charged Ring

You should practice calculating the electrostatic potential \(V(\vec{r})\) due to some simple distributions of charge, especially those with a high degree of symmetry.

Activity 9.6.1. Electrostatic Potential from a Uniform Ring of Charge.

Find the electrostatic potential everywhere in space for a uniformly charged ring with total charge \(Q\) and radius \(R\text{.}\)

Hint.

As you find an integral expression for the electrostatic potential due to a uniform ring of charge, some things you may have needed to pay attention to are:

  1. Draw a picture!

  2. How thick is the ring? What shape is its cross section? Should it be described by a volume charge density? In the absence of further information, the only thing you can do is idealize the ring as a line charge.

  3. In equation (9.5.2), how should the differential volume element \(d\tau'\) be changed to accommodate the fact that the charge density represents a line charge?

  4. In equation (9.5.2), the charge density is described as a volume charge density. You must change this equation to accommodate a line charge.

  5. The information that you are given is the total charge on the ring \(Q\text{.}\) You will need to rewrite the linear charge density \(\lambda\) in terms of the total charge.

  6. Where should the origin be? What coordinate system should you choose?

  7. How can you simplify the differential element in your chosen coordinate system?

  8. How can you simplify the expression for \(|\rr-\rrp|\) in your chosen coordinate system?

Answer.

When evaluated for a ring of charge, the equation (9.5.2) becomes

\begin{gather*} V(r,\phi,z) = \frac{1}{4\pi\epsilon_0} \frac{Q}{2\pi R} \Int_0^{2\pi} \frac{R\,d\phi'}{\sqrt{s^2+R^2-2sR\cos(\phi-\phi')+z^2}} \end{gather*}

Of course, you can combine the various constants in this expression. I chose not to so that you have a better chance of seeing where they came from.

The integral that you found in this activity is, in general, an elliptic integral. It cannot be evaluated in closed form other than to say that its value is the value of that integral. You can look up the value in a table (the old-fashioned way) or get a computer algebra package to compute the value for you (the modern way), but in either case, you do not get an answer which you can easily vary from point to point.

Figure 9.6.1. The level curves of the electrostatic potential \(V\) due to a ring of charge, shown in a vertical planar cross-section with the \(z\)-axis along the left edge of the graph.

To get a better feel for the value of the integral, we consider some limiting cases.

Activity 9.6.2. Limiting Cases of Potential from a Ring of Charge.

Determine the power series expansions that represent the electrostatic potential due to the charged ring, both on axis and in the plane of the ring, and both near to the origin and far from the ring.

Solution.

First of all, let's determine the potential along the axis. In this case, \(\rr\) and \(\rrp\) form the legs of a right triangle, with lengths \(|\rr|=|z|\) and \(|\rrp|=R\text{,}\) respectively. This implies that

\begin{gather*} |\rr-\rrp| = \sqrt{R^2+z^2} \end{gather*}

which brings the integral to the form

\begin{align*} V(z) \amp= \frac{1}{4\pi\epsilon_0} \frac{Q}{2\pi R} \Int_0^{2\pi} \frac{R\,d\phi'}{\sqrt{R^2 + z^2}}\\ \amp= {1\over 4\pi\epsilon_0} {Q\over\sqrt{R^2 + z^2}} \end{align*}

It is instructive to consider the limiting case when \(z=0\text{,}\) which yields

\begin{gather} V(0) = {1\over 4\pi\epsilon_0} {Q\over R}\tag{9.6.1} \end{gather}

which is the potential due to a point charge.

Question 9.6.3. Sensemaking.

Why might you have expected this answer?

Answer.

Since all of the charge is a distance \(R\) from the origin, where you are evaluating the potential, you cannot tell the difference between the potential from a point charge at a distance \(r\) from the ring and the potential at the origin from a ring of charge of radius \(r\text{!}\) Sketch a diagram to see the geometry.

What about the limit as \(z\to0\text{?}\) If \(z\ll R\text{,}\) we have

\begin{align*} {1\over\sqrt{R^2 + z^2}} \amp= {1\over R\sqrt{1+{z^2\over R^2}}}\\ \amp= {1\over R} \left( 1 - {1\over2} {z^2\over R^2} + {3\over8} {z^4\over R^4} - ... \right) \end{align*}

and we have

\begin{gather*} V(z) = {Q\over 4\pi\epsilon_0} \left( {1\over R} - {1\over2} {z^2\over R^3} + {3\over8} {z^4\over R^5} - ... \right) \end{gather*}

The first term, of course, agrees with our previous calculation, Equation (9.6.1) for \(z=0\text{.}\)

Consider the first two terms of this expansion for small \(z\text{.}\) There is no linear term, which means that the potential has a max or a min at \(z=0\text{;}\) this is an equilibrium point. Including the quadratic term results in a parabola — which opens up or down depending on the sign of \(Q\text{.}\) Choosing \(Q>0\) corresponds to a situation with like charges, and like charges repel; a positively charged particle placed along the \(z\)-axis near the origin will be repelled away from the plane of the ring. The origin is therefore an unstable equilibrium point; the parabola opens downwards. Conversely, \(Q\lt 0\) corresponds to opposite charges, which attract; such a particle would be attracted towards the origin, and would oscillate back and forth through the origin. In this case, the origin is a stable equilibrium point; the parabola opens upwards.

Another interesting limit is \(\vert z\vert\to\infty\text{;}\) what happens then? If \(\vert z\vert\gg R\text{,}\) then

\begin{align*} {1\over\sqrt{R^2 + z^2}} \amp= {1\over \vert z\vert} {1\over\sqrt{1+{R^2\over z^2}}}\\ \amp= {1\over \vert z\vert} \left( 1 - {1\over2} {R^2\over z^2} + {3\over8} {R^4\over z^4} - ... \right) \end{align*}

so that

\begin{gather*} V(z) = {Q\over 4\pi\epsilon_0} {1\over \vert z\vert} \left( 1 - {1\over2} {R^2\over z^2} + {3\over8} {R^4\over z^4} - ... \right) \end{gather*}

The leading term is now

\begin{gather*} V(z) \approx {1\over 4\pi\epsilon_0} {Q\over \vert z\vert} \end{gather*}

which is just the potential due to the total charge; to this order, the ring structure is not apparent. Furthermore, the next order term, \(z^{-2}\text{,}\) is zero; i.e. there is no dipole term, which makes sense: since the ring is symmetric as seen from the \(z\)-axis, there is no preferred direction for the dipole moment to point.