Potential V and E field of 2 rings of charge on x axis

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?

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

Re: Potential V and E field of 2 rings of charge on x axis

Quote:

Originally Posted by

**mathnerd15** 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?

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

First let's clean up the notation a bit. The radius of your rings and distance between rings is R not r, according to your problem statement. And the permittivity of free space has a subscript:0

$\displaystyle E = \frac{\lambda 2\pi R x}{4\pi \epsilon _0 (R^2+x^2)^{3/2}}$

$\displaystyle E = \frac{q x}{4\pi \epsilon _0 (R^2+x^2)^{3/2}}$

Now, this is the equation for the electric field of a ring of charge in the yz plane centered at the origin. We have a choice: to place the second ring at x = R or to center the rings at x = +/- R/2. I'm going to go with the more symmetrical +/-R. I'm also going to assume that both rings have the same signed charge. That is to say I'm assuming both rings have charge +q.

The field of the ring at x = -R/2 will be the ring of charge equation translated by R/2:

$\displaystyle E_ = \frac{q \left ( x + \frac{R}{2} \right )}{4\pi \epsilon _0 (R^2 + \left ( x + \frac{R}{2} \right ) ^2)^{3/2}}$

Likewise for the ring at x = +R/2:

$\displaystyle E+ = \frac{q \left ( x - \frac{R}{2} \right )}{4\pi \epsilon _0 (R^2 - \left ( x + \frac{R}{2} \right ) ^2)^{3/2}}$

So do you put these together? (Remember that E is a *vector*.)

-Dan