Problem 3

3)
The easiest two points to consider are the two intercepts with the $\hat{x}$ axis. Call the positive intercept a and the negative intercept b. Then label the distance from q₁ to a to be d. The distance from q₂ to a is then 8.6 nm -d.
So, the potential at a is given by
$V = 0 ~=~ \frac{k q_{1}}{d} + \frac{k q_{2}}{8.6 - d}$
Dividing out k and substituting in for q₁ and q₂, we have
$\frac{6e}{d} = -\frac{-10 e}{8.6 nm - d}$
$d = \frac{3}{8} (8.6 nm)$
d = 3.23 nm
Now, let's look at the potential at b. If the radius of the circle of zero potential is R, then
$V = 0 ~=~ \frac{k q_{1}}{2R - d} + \frac{k q_{2}}{2R - d + 8.6}$
Simplifying yields
$\frac{6e}{2R - 3.23} = -\frac{-10 e}{2R + 5.37}$
Solving for R, we find that
R = 8.06 nm
Now, the center of the circle is located a distance of d - R from the origin.
x_c = d - R
x_c = (3.23 - 8.06) nm
x_c = -4.83 nm