Griffiths confusion

1. Jul 12, 2005

rachmaninoff

I've been going back over Griffiths' E&M this summer, and this question (2.47, electrostatics) is making me stupid:

There are two 'infinite' wires, parallel to each other, with linear charge densities $+ \lambda$ and $- \lambda$;
and the question asks to show that the equipotential surfaces are "circular cylinders".

I don't see how any cylinders could be equipotentials, and in doing the algebra,

(lines parallel to the x axis, + charged one going through (0,d,0), - charged one going through (0,-d,0)

$$V(x,y,z)= -\frac{\lambda}{2 \pi \epsilon _0} \left[ \log \sqrt{ (y-d)^2 + z^2 } - \log \sqrt{ (y+d)^2 + z^2 } \right]$$
\begin{align*}V(r)=V_0 &\Longrightarrow \frac{(y-d)^2+z^2}{(y+d)^2+z^2}=e^{-\frac{4 \pi \epsilon_0}{\lambda} V_0}=C>0 \mbox{ (absolute value signs go away, both the top and bottom are everywhere > 0) } \\ &\Longrightarrow (y-d)^2+z^2 = C\left( (y+d)^2+z^2 \right) \\ &\Longrightarrow y^2+z^2+d^2-2yd=C\left(y^2+z^2+d^2+2yd \right) \\ &\Longrightarrow (1-C)y^2+(1-C)z^2=-(1-C)d^2+4Cyd \end{align}

or,
\begin{align*}r^2&=-d^2+\left( \frac{4C}{1-C}\right) yd, (C > 0) \\ &= d(ky-d), k \in (-\infty,4) \cup (0, + \infty) \end{align}

Which seems to make physical sense - the equipotentials are a family of curves which include y=0 (where V=0!) and curves that look sort of like hyperbolas (all extending to infinity). There are no 'cylindrical' solutions.

Where am I going wrong?

Last edited by a moderator: Jul 12, 2005
2. Jul 12, 2005

rachmaninoff

Edit: my last simplification left out the important C = 1 case, the more general equation was (for the equipotentials):

$$(1-C)y^2+(1-C)z^2=-(1-C)d^2+4Cyd, (C > 0)$$

(from my above post).

That's where you get the equipotential plane y=0: set C=1.

3. Jul 13, 2005

CarlB

Dear rachmaninoff,

The formula for a cylinder centered on the x-axis is:

$$\kappa y^2 + z^2 = \lambda$$

where $$\kappa, \lambda$$ are constants. When you offset such an equation by arranging for the cylinder to be centered on a line going through $$(0,\alpha,0)$$, that is, a line parallel to the x-axis but offset by some amount in the y-axis, the formula becomes:

$$\kappa (y-\alpha)^2 + z^2 = \lambda$$

I think that when you compare this with your formula:

$$(1-C)y^2+(1-C)z^2=-(1-C)d^2+4Cyd$$

you will see that the book is right about it being the equation for a cylinder.

Carl

4. Jul 13, 2005

rachmaninoff

I'm not sure what you mean, there are two charged wires involved (not one), with opposite charges, parallel and with distance 2d between them, and the potential due to each one is

$$V(y,z)=-\int_{O}^{(x,y)} \frac{1}{2 \pi \epsilon_0} \frac{\pm \lambda}{\sqrt{(y\pm d)^2+z^2}}dl$$
$$=\mp\frac{ \lambda}{2 \pi \epsilon} \log | \sqrt{(y\pm d)^2+z^2} | +V_0$$

My point is that when you superimpose the two potentials, you don't get any equipotential cylinders, which is why my 'formula' is different from the book formula for a cylinder. That's my problem.

Last edited by a moderator: Jul 13, 2005
5. Jul 13, 2005

CarlB

Your problem is that you were just wrong in this statement. The curves don't look at all like hyperbolas. They're cylinders. The book is right, you're wrong. This is the normal order of the universe. Don't feel bad, it's not all that obvious.

The equation defining the surface that you gave was correct:

$$y^2 + z^2 =-d^2+\left( \frac{4C}{1-C}\right) yd, (C > 0)$$

Let me rewrite it for you so that it becomes more obvious that this is a cylinder. Complete the square for y:

$$\left(y-\frac{2dC}{1-C}\right)^2 + z^2 = d^2\left(\frac{4C^2}{(1-C)^2}-1\right) = \Delta^2$$

This is the equation for a cylinder with center $$(0,2dC/(1-C),0)$$ extending in the x direction, and with a diameter of $$\Delta$$ . Note that the cylinder is not centered on either of the two wires (as were those wires' equipotential surfaces in the absence of the other wire). Instead, the cylinders get pushed away from the other wire.

If you still have doubts about this, you need to fix it because this is really very basic electrostatics. Here's an applet that will help you visualize the situation:

In the above, click on "dipole charge" to see the way the cylinders look.

Now think carefully about what you are saying, that is, that the equipotential surfaces extend to infinity. Do you really think that it is physically realistic for a pair of wires (whose charge adds to zero) to have a field that extends to infinity with an undiminished intensity?

Carl

Last edited: Jul 13, 2005
6. Jul 13, 2005

rachmaninoff

Thanks, I got it!