# Point of equilibrium with 2 charged spheres

## Homework Statement

Two arbitrarily thin concentric coplanar rings of radius a and 2a carry uniformly distributed charge density -Qe and $$\sqrt{27}$$Qe, respectively.
Assuming that Z axis coincides with the rings' axes, and that the plane of the rings correspond to z=0, determine the location of all points of equilibrium(ie zero E field).

## The Attempt at a Solution

Need some hint.

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gabbagabbahey
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## The Attempt at a Solution

Need some hint.
Use the superposition principle.

superposition came to my mind too. But didn't get any further. I consider the first sphere with charge -Qe. How find where E=0.

gabbagabbahey
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Sphere? What sphere?

Your problem statement tells you you have two infinitesimally thin rings of charge, not spheres.

Anyways, the superposition principle allows you to calculate the total electric field by adding the electric fields of each ring together. So, a good place to start might be to actually calculate the electric field of each ring.

Well, I used gauss' law and calculated the E field produced by each ring.

E1 = -KQZ/ (Z2+a2)3/2
and
E2 = KQ(27)1/2Z/ (Z2+4a2)3/2

I tried to find point where E1+E2=0 along Z. I just got lost in math. Am I going in the right direction?

gabbagabbahey
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Well, [I used gauss' law and calculated the E field produced by each ring.

E1 = -KQZ/ (Z2+a2)3/2
and
E2 = KQ(27)1/2Z/ (Z2+4a2)3/2

I tried to find point where E1+E2=0 along Z. I just got lost in math. Am I going in the right direction?
Do you mean Coulomb's Law? There are none of the requisite symmetries present in this problem for Gauss' Law to be directly applicable.

Anyways, the fields you've given only give the z-components of each field (Remember, electric fields are vectors... they have 3 components and all 3 must be zero for equilibrium points) and only in the very special case that the field point is on the axis of the rings...what about points that are off-axis?

As for finding on-axis points where E=0; to make the math easier you might want to notice that $\sqrt{27}=3^{3/2}$. So, if you end up with an equation like $\sqrt{27}(z^2+a^2)^{3/2}-(z^2+4a^2)^{3/2}=0$ (which you should), You could rewrite it as $(3z^2+3a^2)^{3/2}=(z^2+4a^2)^{3/2}$. Squaring both sides of the equation will give you a 6th degree polynomial to find the roots of and hence there will be 6 roots (although they may not all be real valued). You should immediately notice however that two of those roots will satisfy the equation $(3z^2+3a^2)=(z^2+4a^2)$ which you should have no difficulty solving.

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