# Cassini ovals being the equipotential lines?

1. Oct 5, 2011

### torteloni

edit: If you cannot understand a passage it is probably because of my bad English. Just ask.

Dear board,

it is said oftentimes that the cassini ovals are the equipotential lines when considering two (equal) point charges in the oval's foci. (see e.g. http://www.hst.tu-darmstadt.de/uploads/media/hvt2_v_08b.pdf [Broken] )

I now wonder how this can be shown. Here is what I tried and what however did not work out the way I hoped. I wanted to show that if the product of distances to two given point charges is the same for some points in space, the potential in these points is the same:

Consider two equally charged point charges Q1 and Q2. Consider any point P in space. Let r1 and r2 be the distances between that point and Q1 and Q2 respectively. The electric potential in P created by Q1 and Q2 is:

$\frac{1}{4 \pi \epsilon _0} \cdot \frac{Q_1}{r_1} + \frac{1}{4 \pi \epsilon _0} \cdot \frac{Q_2}{r_2}$

$= \frac{1}{4 \pi \epsilon _0} \cdot (\frac{Q_1}{r_1} + \frac{Q_2}{r_2})$

We just write $K$ instead of the constant $\frac{1}{4 \pi \epsilon _0}$ from now on. Besides we define $Q:=Q_1=Q_2$ ($Q_1=Q_2$ by our initial assumption). So we receive:

$= K \cdot (\frac{Q}{r_1} + \frac{Q}{r_2})$

$= K \cdot Q \cdot (\frac{1}{r_1} + \frac{1}{r_2})$

$= K \cdot Q \cdot \frac{r_1+r_2}{r_1\cdot r_2}$

And here at this point I don't quite get it: Even if the product of distances remains constant, the potential energy may change.

Consider a point in space with $r_1=4$ and $r_2=1$. The potential in this point is obviously $K\cdot Q\cdot \frac{5}{4}$.

Now consider a point with $r_1=2$ and $r_2=2$. Because $4\cdot 1=2\cdot 2$ both points lie on the same cassini oval around $Q_1$ and $Q_2$. However in our second point we have the potential $K\cdot Q\cdot \frac{4}{4}$.

Where am I mistaking? I thank you very much for any help and I'm especially glad if you can show me helpful links or literature that deal with this topic.

Last edited by a moderator: May 5, 2017
2. Dec 23, 2011

### torteloni

By now I have found my error. My calculation for point charges is correct, however the Cassini Ovals are the lines of equal potential regarding a logarithmic potential. So $Q_1$ and $Q_2$ are not point charges but electrically charged conductors.

The electrostatic potential for one point P in the plane is then:

$V=- \frac{q}{2 \pi \epsilon _0} (ln(r)+ln(r'))$

where r and r' are the distances to the two electrally charged conductors respectively. You may want to check http://www.jstor.org/stable/3620950 for further information.