- #1

Angela G

- 65

- 19

- Homework Statement
- Consider a circular ring with radius R and uniform longitudinal charge density λ

a) Assume that a charge of magnitude q is in the center of the circle. The power of it then becomes zero, i.e. we have an equilibrium position. Is this stable or unstable? Motivate the answer!

b) Suppose now that we take off half of the ring and that the long charge density remains the same. (The ring is not conductive.) Determine the magnitude of the force acting on the charge q.

- Relevant Equations
- ## V(0) = \frac{\lambda}{2\epsilon_0} ##

## \nabla^2 V = 0 ##

## U = - \vec p \cdot \vec E ##

## W = q V ##

I wonder if you could help me with both I'm stuck, I know that in order to see if the electrostatic equilibrium is stable or not at the center of the ring , the potential energy has to be minimum there. I was going to use Laplace eq. but it allows neither minimum nor maximum. Then I also thought that if you move the point charge to either the right or left, it should experience a force that takes the point charge back to the center of the ring (for stable equilibrium). But since we have only got the magnitude of charges, I guess I have to make assumptions (if both the point and line circles have the same sign or if they have different signs of charges). Another alternative that I thought of was to use that U = - p * E, where if the electric dipole is parallel to the electric field we have the smallest potential energy, but again the problems arise about the signs of the charges and now it is also an external electric field involved, which we do not have. So I'm wondering if you could guide me. I would like to solve in with advanced physics, please.

I determined the electric potential at the center of the ring without the charge it was $$ V(0) = \frac{\lambda}{2\epsilon_0} $$

for b) I was thinking that I can determine the electric field cause we don't hade longer circular simetry and then the force by columbs law

I determined the electric potential at the center of the ring without the charge it was $$ V(0) = \frac{\lambda}{2\epsilon_0} $$

for b) I was thinking that I can determine the electric field cause we don't hade longer circular simetry and then the force by columbs law