# Homework Help: SOLVED: Equipotential surfaces for finite line of charge

1. Apr 22, 2015

### homer

1. The problem statement, all variables and given/known data
Purcell 2.10
A thin rod extends along the z axis from $z = -d$ to $z = d$. The rod carries a charge uniformly distributed along its length with linear charge density $\lambda$. By integrating over this charge distribution calculate the potential at a point $P_1$ on the $z$ axis with coordinates $0,0,2d$. By another integration find the potential at a point $P_2$ on the $x$ axis and locate this point to make the potential equal to the potential at $P_1$

The points $P_1$ and $P_2$ in the preceding problem happen to lie on an ellipse which has the ends of the rod as its foci, as you can readily verify by comparing the sums of the distances from $P_1$ and from $P_2$ to the ends of the rod. This suggests that the whole ellipse might be an equipotential. Test that conjectire by calculating the potential at the point $(3d/2, 0, d)$ which lies on the same ellipse. Indeed it is true, though there is no obvious reason why it should be, that the equipotential surfaces of this system are a family of confocal prolate spheroids. See if you can prove that. You will have to derive a formula for the potential at a general point $(x,0,z)$ in the $xz$ plane. Then show that, if $x$ and $z$ are related by the equation
$$\frac{x^2}{a^2 - d^2} + \frac{z^2}{a^2} = 1,$$
which is the equation for an ellipse with foci at $z = \pm d$, the potential will depend only on the parameter $a$, not on $x$ or $z$.

2. Relevant equations
Setting our zero of potential at infinity, the potential of a point charge $q$ at a point a distance $r$ away is
$$\varphi = \frac{q}{r}.$$
Note we're using cgs units, not SI units.

The equation of an ellipse with foci $z = \pm d$ and semimajor axis $a$ along the $z$ axis and semiminor axis $a^2 - d^2$ along the $x$ axis in the $xz$ plane
$$\frac{x^2}{a^2 - d^2} + \frac{z^2}{a^2} = 1.$$

3. The attempt at a solution
Consider a small charge element $dq$ of length $dz'$ of the rod, at position $z'$ on the $z$ axis. Treating this small charge element as a point charge, the infinitesimal potential at $(x,0,z)$ due to this charge element is
$d\varphi(x,0,z) = \dfrac{dq}{\sqrt{x^2 + (z - z')^2}} = \dfrac{\lambda\,dz'}{\sqrt{x^2 + (z-z')^2}}.$
Integrating along the entire rod from $z = -d$ to $z = +d$. we find the potential to be given by the integral
$$\varphi(x,0,z) = \int_{-d}^d \frac{\lambda\,dz'}{\sqrt{x^2 + (z-z')^2}}$$

If we make the trig substitution $z - z' = x\tan{\theta}$, we get $-dz' = x\sec^2{\theta}\,d\theta$ and our integral becomes
\begin{align*} \varphi(x,0,z) & = \int_{\theta_1}^{\theta_2} \frac{-\lambda\, x\sec^2{\theta}\,d\theta}{\sqrt{x^2(1 + \tan^2{\theta})}} \\ & = \int_{\theta_2}^{\theta_1} \frac{\lambda\, x\sec^2{\theta}\,d\theta}{\sqrt{x^2 \sec^2{\theta}}} \\ & = \int_{\theta_2}^{\theta_1} \lambda\,\sec{\theta}\,d\theta \\ & = \lambda\, \text{ln}\Big[ \frac{\sec{\theta_1} + \tan{\theta_1}}{\sec{\theta_2} + \tan{\theta_2}} \Big], \end{align*}
where $z - d = x\tan{\theta_2}$ and $z + d = x\tan{\theta_1}$, so that
\begin{align*} \sec{\theta_1} & = \frac{\sqrt{x^2 + (z + d)^2}}{x} \\ \sec{\theta_2} & = \frac{\sqrt{x^2 + (z-d)^2}}{x}. \end{align*}
Then our integral becomes
\begin{align*} \varphi(x,0,z) & = \lambda\, \text{ln}\Big[ \frac{\frac{1}{x} \sqrt{x^2 + (z+d)^2} + \frac{1}{x}(z+d)} {\frac{1}{x} \sqrt{x^2 + (z-d)^2} + \frac{1}{x}(z-d)} \Big] \\ & = \lambda\, \text{ln}\Big[ \frac{z + d + \sqrt{x^2 + (z+d)^2}} {z - d + \sqrt{x^2 + (z-d)^2}} \Big]. \end{align*}

Now from the ellipse equation we have
$$x^2 = (a^2 - d^2) \Big(1 - \frac{z^2}{a^2}\Big),$$
and we thus find
\begin{align*} x^2 + (z \pm d)^2 & = (a^2 - d^2)\Big(1 - \frac{z^2}{a^2}\Big) + (z \pm d)^2 \\ & = \Big(a^2 - d^2 - z^2 + \frac{d^2}{a^2}z^2\Big) + (z^2 \pm 2d z + d^2) \\ & = a^2 + \frac{d^2}{a^2}z^2 \pm 2d z \\ & = \frac{1}{a^2}\Big(d^2 z^2 \pm 2 a^2 d z + a^4\Big) \\ & = \frac{1}{a^2}(d z \pm a^2)^2 \end{align*}
Thus we find
$$\sqrt{x^2 + (z \pm d)^2} = \frac{1}{a} \lvert d z \pm a^2\rvert$$
so that our expression for the potential at point $(x,0,z)$ becomes
\begin{align*} \varphi(x,0,z) & = \text{ln}\Big[ \frac{z + d + \lvert dz + a^2\rvert/a}{z - d + \lvert dz - a^2\rvert/a} \Big] \\ & = \text{ln}\Big[ \frac{a z + a d + \lvert dz + a^2\rvert}{a z - a d + \lvert dz - a^2\rvert} \Big] \end{align*}
Now I'm lost as to how to get rid of the $z$ term so that
$$\varphi(x,0,z) = F(a,d)$$
for some function $F$ independent of $x,z$, as the problem requests us to find.

Last edited: Apr 22, 2015
2. Apr 22, 2015

### homer

Let's just for the minute assume $x > 0$ also.

3. Apr 22, 2015

### homer

I think I figured it out now. Always seems to happen right after I type up the whole problem here!

4. Apr 22, 2015

### homer

By the symmetry of the problem sign of $x,z$ doesn't matter, so I'll just take both positive. Then $\lvert dz - a^2\vert = a^2 - dz$ and $\lvert dz + a^2\rvert = dz + a^2$, so the expression for the potential becomes
\begin{align*} \varphi(x,0,z) & = \lambda\,\text{ln}\Big[ \frac{az + ad + a^2 + dz}{az - ad + a^2 - dz} \Big] \\ & = \lambda\,\text{ln}\Big[ \frac{a(z + a) + d(z+a)}{a(z+a) - d(z+a)} \Big] \\ & = \lambda\,\text{ln}\Big[\frac{a+d}{a-d}\Big]. \end{align*}

Now I just have to check the separate cases $x = 0, z = a$ and $x = \sqrt{a^2 - d^2}, z = 0$.

5. Apr 22, 2015

### homer

This was a cool problem. Man I love Purcell.