there are 2 rings of charge, radius R on the x axis separated by a distance R, find the potential and E field.

so don't you have to calculate the field to the left of the 2 rings, in between the 2 rings and to the right? I get these answers for the E field which by symmetry points only on the x axis away from the rings. in the area to the left and to the right of the rings the 2 fields will add and in between there will be some subtraction

I think my calculation is off here, the field for one ring, positive lambda charge pointing away from the ring is?

$\displaystyle \frac{\lambda 2\pi rx}{4\pi \epsilon (r^2+x^2)^{3/2}}$

$\displaystyle \frac{-2qx-qr}{4\pi \epsilon (r^2+x^2)^{3/2}}, \frac{q(2x-r)}{4\pi \epsilon (r^2+x^2)^{3/2}}, \frac{2qx+qr}{4\pi \epsilon (r^2+x^2)^{3/2}}$

calculate only the field on the x axis, the absolute value of the fields to the left and to the right of 2 rings are equal

for V to the right of 2 rings:

$\displaystyle -\int_{\infty }^{x}\frac{2qx+qr}{4\pi \epsilon (r^2+x^2)^{3/2}}= \frac{q}{4\pi \epsilon r(\frac{1}{r^2}+1)}$

for V to the left of the 2 rings, is the potential thus, or can I use a potential from -infinity to x3 the point intersecting the left ring?

$\displaystyle -\int_{\infty }^{x}\frac{2qx+qr}{4\pi \epsilon (r^2+x^2)^{3/2}}- \int_{x1}^{x2}E2dl-\int_{x2}^{x3}E3dl= \frac{q}{4\pi \epsilon r(\frac{1}{r^2}+1)}-V2-V3$