Consider a collection of three charges arranged in a line along the \(z\)-axis: charges \(+Q\) at \(z=\pm D\) and charge \(-2Q\) at \(z=0\).
Find the electrostatic potential at a point \(\vec{r}\) in the \(xy\)-plane at a distance \(s\) from the center of the quadrupole. The formula for the electrostatic potential \(V\) at a point \(\vec{r}\) due to a charge \(Q\) at the point \(\vec{r}'\) is given by: \[ V(\vec{r})=\frac{1}{4\pi\epsilon_0} \frac{Q}{\vert \vec{r}-\vec{r}'\vert} \]
We will choose to work in cylindrical coordinates, and because we want a point in the \(xy\)-plane, we know \(z=0\), so our vector that points to where we care about measuring the potential in space, \(\vec{r}\), becomes \(\vec{r}=s \hat{s}+z \hat{z}\rightarrow s\hat{s}\).
Electrostatic potentials satisfy the superposition principle. By the principle of superposition, the potential due to the three charges is just the sum of the potentials due to the individual charges, so \(V_{total}=V_1+V_2+V_3\). Since they're all point charges, we can use the given equation for a point charge for each individual potential, \(V(\vec{r})=\frac{1}{4\pi \epsilon_0}\frac{Q}{\left|\vec r-\vec r^{\prime}\right|}\). The only difference will be the charges of each contributing potential, \(Q\), and the vectors we will use that point from the origin to the charges themselves, \(\vec{r}'\).
So, we can write the total potential out like so: \begin{align} V(\vec{r})&=V_1(\vec{r})+V_2(\vec{r})+V_3(\vec{r})\\ &= \frac{1}{ {4\pi\epsilon_0}}\left\{\frac{Q_1}{\left|\vec r-\vec r^{\prime}_1\right|} +\frac{Q_2 }{ \left|\vec r-\vec r^{\prime}_2\right|} + \frac{Q_3}{\left|\vec r-\vec r^{\prime}_3\right|}\right\} \end{align} Now we just make an arbitrary decision about which number corresponds to which charge, and we let charges \(Q_1=+Q\) and be located at \(z=+D\), so that \(\vec{r}_{1}=D\hat{z}\). Similarly, \(Q_3=+Q\) and be located at \(z=-D\), so that \(\vec{r}_3=-D\hat{z}\), and finally, \(Q_2=-2Q\) and be located at \(z=0\), so that \(\vec{r}_2=(0)\hat{z}=0\). Plugging these into our whole equation for \(V(\vec{r})\), we get: \begin{align} V(\vec{r}) &= \frac{1}{{4\pi\epsilon_0}} \left\{\frac{Q}{\left|s \hat s -D\hat z\right|} -\frac{2Q}{s} + \frac{Q}{\left|s \hat s+D\hat z\right|}\right\} \end{align} Doing vector dot products to get the magnitude, we get a really fun result due to the symmetry of our problem (and the fact we're only looking on the xy-plane): \begin{align} \left|s \hat s+D\hat z\right|=\left|s \hat s-D\hat z\right|=\sqrt{s^2+D^2} \end{align} Both these magnitudes ended up being the same! And we plug this back in and we're able to simplify our potential to: \begin{align} V(\vec{r}) &= \frac{2Q}{4\pi\epsilon_0}\left\{\frac{1}{\sqrt{s^2 + D^2}} -\frac{1}{s}\right\} \end{align} Sense-making: These equations should be starting to get familiar, but checking the dimensions should remain a go-to strategy. The constants out front carry most of the dimensions, with inverse distance left---and sure enough, each term manages to have dimensions of inverse distance. Another great way of sense-making is to graph this thing and see what it looks like, which should also become one of your favorite strategies!
Assume \(s\gg D\). Find the first two non-zero terms of a power series expansion to the electrostatic potential you found in the first part of this problem.
In the first term from your answer to part (a) above, we recognize we need to use the memorized series expansion for \((1+z)^p\), which is: \begin{align} (1+z)^p= 1+pz+\frac{p(p-1)}{2!}z^2+\frac{p(p-1)(p-2)}{3!}z^3\dots \end{align} Now we can manipulate the expression by pulling out a factor of \(s^2\) from the square root: \begin{align} V(\vec r) &= \frac{2Q}{4\pi\epsilon_0}\left\{\frac{1}{\sqrt{s^2 + D^2}} -\frac{1}{s}\right\}\\ &= \frac{2Q}{4\pi\epsilon_0} \left\{\frac{1}{s\sqrt{1 + \left(\frac{D}{s}\right)^2}} -\frac{1}{s}\right\}\\ &= \frac{2Q}{4\pi\epsilon_0} \left\{\frac{1}{s} \left(1+\left(\frac{D}{s}\right)^2\right)^{-\frac{1}{2}} -\frac{1}{s}\right\} \end{align} Now it looks like \((1+z)^p\), with \(z=\left(\frac{D}{s}\right)^2\) and \(p=-\frac{1}{2}\), so we expand that term: \begin{align} V(\vec r) &= \frac{2Q}{4\pi\epsilon_0} \left\{\frac{1}{s}\left(1 -\frac{1}{2} \left(\frac{D}{s}\right)^2 +\frac{3}{8}\left(\frac{D}{s}\right)^4+\dots\right) -\frac{1}{s}\right\} \end{align} Now we distribute in the \(\frac{1}{s}\) and notice the first term cancels with the \(-\frac{1}{s}\) term: \begin{align} V(\vec r) &= \frac{2Q}{4\pi\epsilon_0} \left\{-\frac{1}{2}\frac{D^2}{s^3} +\frac{3}{8}\frac{D^4}{s^5}+\dots\right\} \end{align}
This charge distribution is called a quadrupole because the total charge and the dipole moment are both zero. In your equation, that actually means that terms in the potential that go like \(1/s\) (the "monopole" term) and \(1/s^2\) (the "dipole" term) are zero. (You have only calculated these quantities on the \(xy\)-plane, but it turns out that they are zero everywhere.) Therefore, the first (dominant) non-zero term in what's called a "multipole" expansion is the \(1/s^3\) (the "quadrupole" term). The charge distribution is called a linear quadrupole because the charges all lie in a straight line. Can you think of another (nonlinear) distribution of charges that could also be called a quadrupole? Consider a series of four charges arranged in a square with equal positive and negative charges alternating around the square.
In this problem, in all cases, you are expected to evaluate any integrals in your answers.
Along the axis of symmetry, \(s=0\), and the integral becomes \begin{align} V(z) &= \frac{1}{4\pi\epsilon_0} \int_0^{2\pi}\frac{\lambda\,R\,d\phi'}{\sqrt{R^2 + z^2}}\\ &= \frac{1}{4\pi\epsilon_0} \frac{2\pi\lambda\,R}{\sqrt{R^2 + z^2}}\\ &= \frac{1}{4\pi\epsilon_0} \frac{Q}{\sqrt{R^2 + z^2}} \end{align} Sense-making: Always check dimensions on these potential expressions---especially because electric fields are coming up, and they have different dimensions. Also, note that there's an interesting symmetry between \(R\) and \(z\) here: moving a small distance away from the ring along the \(z\)-axis has the same effect as staying put while making the ring a little bit bigger.
The disk is a 2D object, so we need a \(\sigma\) to describe it, thus our tiny amount of charge takes on the form: \begin{equation} \sigma dA = \sigma\, s' \,ds' d\phi' \end{equation} So we use this in our general expression for \(V\): \begin{align} V(z) &= \frac{1}{4\pi\epsilon_0}\int \frac{\sigma(r') dA'}{|\vec{r}-\vec{r}'|} \\ &= \frac{\sigma}{4\pi\epsilon_0}\int_0^{2\pi} \int_0^R \frac{s'\,ds' \, d\phi'}{\sqrt{s'^2 + z^2}} \\ &= \frac{2\pi\sigma}{4\pi\epsilon_0}\int_0^R \frac{ s'\,ds'}{\sqrt{s'^2 + z^2}} \\ &= \frac{2\pi\sigma}{4\pi\epsilon_0} \sqrt{s'^2 + z^2} \Bigg|_0^R \\ &= \frac{2\pi\sigma}{4\pi\epsilon_0} \left( \sqrt{R^2 + z^2} - |z| \right) \end{align}
Sense-making: Pay close attention to the different dimensions of \(\sigma\), and you might even want to put the whole expression back in terms of \(Q\) like we did for the previous part. Also, at first glance it seems like this potential might not be falling off as \(z\) gets larger, which wouldn't make sense at all. You might want to graph the potential, or even graph the two terms in the potential separately, to see that the first term may start off higher, but that it quickly approaches the second term in value, while always remaining positive.
A conical surface (an empty ice-cream cone) carries a uniform charge density \(\sigma\). The height of the cone is \(H\), and the radius of the top is \(R\).
Sense-making: It is interesting to note that the result increases with \(H\), so as I make the cone larger, the potential at this point goes up. This may be initially unintuitive, as the point of interest is getting farther away from where the charge is, but because the cone has charge all the way around it, and the charge density is constant, you actually are accumulating more charge than you need to balance out the increasing distance.
The cone that I'll be considering has its vertex at the origin and opens up on the positive z-axis. I'll use cylindrical coordinates in my solution (but I could also use spherical if I wanted). Starting with the general expression for calculating potential due to a surface charge distribution: \begin{equation} V(\vec{r})=\int \int \frac{k \,\sigma \,dA'}{|\vec{r}-\vec{r}\,'|} \end{equation} then I'll use what I know. For a cone with height \(H\) and base radius \(R\), the equation of the cone is \(z'=\frac{H}{R}\,s'\).
Draw two infinitesimal displacement vectors on the surface of the cone: \begin{equation} d\vec{r}_1' = s'\,d\phi'\,\hat{\phi}' \qquad d\vec{r}_2 '= ds'\,\hat{s}' + dz'\,\hat{z}' = \left(\hat{s}' + \frac{H}{R}\,\hat{z}'\right)\,ds' \end{equation} which in turn yields \begin{equation} dA' = |d\vec{r}_1'\times d\vec{r}_2'| = \sqrt{1+\frac{H^2}{R^2}}\; s'\, ds'\,d\phi' \end{equation}
Using this differential area element, and inserting the expression discussed in class (and in previous HW) for the distance \(|\vec{r}-\vec{r}\,'|\) in cylindrical coordinates, the potential becomes: \begin{equation} V(\vec{r}) = \int_0^{2\pi} \int_0^R \frac{k \,\sigma\sqrt{1+\frac{H^2}{R^2}} \,s'\,ds'\,d\phi'} {\sqrt{s^2+s'^2-2ss'\cos(\phi-\phi') +(z-z')^2}} \end{equation} Since we know where the potential is being evaluated, we can simplify the denominator; we have \(z=H\) and \(s = 0\), so that: \begin{align} V(s=0,z=H) &= \int_0^{2\pi} \int_0^R \frac{k \,\sigma\sqrt{1+\frac{H^2}{R^2}} \,s'\,ds'\,d\phi'} {\sqrt{{s'}^2 + (H-z')^2}} \\ &= \int_0^{2\pi} \int_0^R \frac{k \,\sigma\sqrt{1+\frac{H^2}{R^2}} \,s'\,ds'\,d\phi'} {\sqrt{{s'}^2 + (H-\frac{H}{R}s')^2}} \\ &= k \,\sigma\sqrt{1+\frac{H^2}{R^2}} \,\int_0^{2\pi} \int_0^R \frac{s'\,ds'\,d\phi'} {\sqrt{{s'}^2 + \left(\frac{H}{R}\right)^2(R-s')^2}} \end{align}
- Using technology or an integral table to evaluate the integral, we finally obtain: \begin{equation} V(s=0,z=H) = \pi k\,\sigma H \,\ln{\left[\frac{\sqrt{2}+1}{\sqrt{2}-1}\right]} \end{equation}