Potential of ring with sphere inside it

[adamp121]

Homework Statement

Hi,

I'm trying to find the potential of conducting grounded sphere with radius Rs which located in the center of charged ring with Rr (>Rs) with charge density λ, h meters up to the z axis (see the attached images)


Rs=4.3[cm]
Rr=6.6[cm]
h=13.1[cm]
λ=1.0[esu/cm]

The answer should be 0.81023024

2. The attempt at a solution

The sphere potential is easy one -
\Phi ring=\frac{2\pi Rr\lambda}{\sqrt{z^2+Rr^2}}
But how do I get the potential of the sphere?
I've tried to use the image charged method, but I can't get to the right answer.
Can somebody help me?

Thanks,
Adam

 

[haruspex]

I'm trying to find the potential of conducting grounded sphere

Doesn't sound too hard . Am I missing something?

 

[adamp121]

Can you give me some clue about the calculation of the sphere's potential?

 

[Saitama]

Can you give me some clue about the calculation of the sphere's potential?

I don't know how to calculate the sphere's potential but the problem can be easily done by method of image charges. Did you try that?

 

[adamp121]

Yes.

I know how to replace a grounded sphere near to a single particle, but don't know how to replace a grounded sphere with surrounded ring.
I have only the ring’s potential on the Z axis, and it makes it harder.

 

[Saitama]

I know how to replace a grounded sphere near to a single particle, but don't know how to replace a grounded sphere with surrounded ring.

Like a point charge, you can replace the grounded sphere with a ring here in this case.

 

[adamp121]

I've tried it also, where I set the image ring with charge -
q'=-2/pi\frac{Rs}{Rr}
and
R'=\frac{Rs^2}{Rr}

as described here -
http://en.wikipedia.org/wiki/Method_of_image_charges#Method_of_images_for_spheres

 

[Saitama]

I've tried it also, where I set the image ring with charge -
q'=-2/pi\frac{Rs}{Rr}

Do you mean $\displaystyle q'=-\frac{R_s}{R_r}Q$ where $Q=\lambda \cdot 2\pi R_r$?

R'=\frac{Rs^2}{Rr}

Correct!