Unstable or stable electrostatic equilibrium?

[PeroK]

I looked at what happened if we displace the point charge in the z- axis, we call the displacement dz. So by symmetry all the element charges of the ring will act on the charge q. But they will not longer cancel each other. Know we will have a non- zero field,$dE_z$, acting on a charge at a distance dz above the center of ring . The non-zero field is in the positive z-direction . Then the force, $dF_z$ due to the ring will point away from the ring and thus there is no restoring force on the z-direction . It means that the equilibrium is not statble in the z- direction

That's true if the charges are of the same sign. But, if the charges are of opposite sign, there there is stable equilibrium in the z-direction.

 

[PeroK]

If I wasn't given a sign for the charge I'd assume it was positive.

In general, a line charge of $\lambda$ may be positive or negative; likewise $q$ may be positive or negative.

 

[Angela G]

In general, a line charge of λ may be positive or negative; likewise q may be positive or negative.

exacly, we assumed that both charges are positive

 

[PeroK]

exacly, we assumed that both charges are positive

Which is an unjustified assumption. $\lambda$ and $q$ could have opposite signs.

 

[Angela G]

Which is an unjustified assumption. λ and q could have opposite signs

that's true, how can we know the signs?

 

[PeroK]

that's true, how can we know the signs?

You have to analyse both cases. Same sign and opposite sign.

 

[Angela G]

You have to analyse both cases. Same sign and opposite sign

ok, thanks I'll keep that in mind

 

[PhDeezNutz]

In general, a line charge of $\lambda$ may be positive or negative; likewise $q$ may be positive or negative.

I think we should assume opposite signs now that I think about it.

if it were the same signs the point charge wouldn’t experience a restoring force to the center of the ring when displaced up and down the axis of symmetry. It would be repelled. In which case the problem would be a trivial one liner.

I guess for completeness the OP should address this case.

your thoughts?

 

[PeroK]

I think we should assume opposite signs now that I think about it.

if it were the same signs the point charge wouldn’t experience a restoring force to the center of the ring when displaced up and down the axis of symmetry. It would be repelled. In which case the problem would be a trivial one liner.

I guess for completeness the OP should address this case.

your thoughts?

Yes, in 3D the instability of the system for the same sign can be seen simply. That leaves the opposite sign case to be determined.

 

[Angela G]

ok, so If the charges has oposite sign, the there will be an attractive force when the point charge is displaced a dz from the ring, and know we have to determine the stability on the plane of the ring.

Also, I wonder if you could take a look at post #50, and tell me if my reasoning is right or not, pls

 

[PeroK]

ok, so If the charges has oposite sign, the there will be an attractive force when the point charge is displaced a dz from the ring, and know we have to determine the stability on the plane of the ring.

Also, I wonder if you could take a look at post #50, and tell me if my reasoning is right or not, pls

I don't think you were supposed to get into detailed calculations. That's why the question said simply "motivate your answer". You're looking for a good approximation for the force on a particle that is perturbed off centre.

 

[kuruman]

ok, so If the charges has oposite sign, the there will be an attractive force when the point charge is displaced a dz from the ring, and know we have to determine the stability on the plane of the ring.

You have examined $$- \frac{\partial^2 V}{\partial z^2} = \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial V}{\partial \rho} \right )$$ This says that:
1. If one side of the equation is positive (it doesn't matter which) the other is negative.
2. The positive side means stable equilibrium in that direction and the negative side means unstable equilibrium in that direction.

Is that observation going to change if the signs of the charged ring and the test particle are the same as opposed to different?

This thread has gone on for too long in my opinion. One can find more details about the potential and electric fields near the origin (for like charges) in post #2 here.

 

[bob012345]

ok, so If the charges has oposite sign, the there will be an attractive force when the point charge is displaced a dz from the ring, and know we have to determine the stability on the plane of the ring.

Also, I wonder if you could take a look at post #50, and tell me if my reasoning is right or not, pls

I think you already have said the way the ring field acts and you know which way it makes a charge with a small displacement go for each sign of $q$.
Just spell it out.

It is interesting to compare the stability in the plane vs. out of the plane along the $z$ axis in both cases (of like or opposite charges).

 

[bob012345]

Which is an unjustified assumption. $\lambda$ and $q$ could have opposite signs.

It was used as a starting case, that's all. The possibility of different signs were discussed earlier.

 

[bob012345]

I think part $a$ is basically solved or at least enough was discussed to complete it, so what about part $b$?

 

[Angela G]

could you please summarize a) because I'm little confused

 

[bob012345]

could you please summarize a) because I'm little confused

You solved the stability for the $z$ axis above and you know the direction and therefore force in the plane (toward the center for positive ring) and so you know whether it is stable in the plane of the ring given the sign of the charges. If you want to be more detailed for example on the $z$ axis just compute the potential along the $z$ axis which is easy because of symmetry and take the second derivative of it.

 

[Angela G]

regarding to b) I think I already solved it could you please take a look of it

Edit: I missed a "q" in the third expression

 

[PeroK]

regarding to b) I think I already solved it could you please take a look of it

Edit: I missed a "q" in the third expression

Looks correct.

 

[Angela G]

Thank you everyone that helped me in this problem I finally understood