# Method of images

1. Jan 13, 2007

### stunner5000pt

1. The problem statement, all variables and given/known data
An infinite conducting sheet has a hemispherical bubble of radius a. (Refer to the diagram)

2. The attempt at a solution

ok i know the boundary conditions should be (Phi refresents the potential)
And we are working in CGS units... units with which i am not entire comfortable.
Z is the vertical and Y is the horizontal
$\Phi(y, \sqrt{a^2 - y^2}) = 0$ for $-a\leq y\leq a$
$\Phi(y,0) = 0$ where y<= -a and y=> a

Should i be considering the 3D case because it does say bubble...
i know how to do this for a plane and a sphere but this 'mixed' case has got me confused.

i thought of locating a charge at a point (-z0+2a,0) and that satisifes condition 1 but not condition 2

it doesnt work for any point other tan this one so no..

maybe i am not thinking about something and so im stuck!

i think that there would be more than one point charge on th segment y = -z0 +2a

Thanks for your input!

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2. Jan 13, 2007

### Allday

hi stunner. i think you should be thinking about the 3d case. i remember doing this problem a while back. your right to set the potential to zero on the conductor. i remember having to place more than one image charge to make it work.

3. Jan 14, 2007

### mjsd

NB: the 3D case is actually much simpler... you may need a series of images .. hopefully it will converge (ie. the magnitude of the image charge you insert will progressively become smaller) my guess is that all of them will be on the Z axis judging by the symmetry of the problem.

4. Jan 16, 2007

### stunner5000pt

i was thinking there were more than 1 image charges

i thought of a sperposition between the case of the plane and the case of a hemisphere (which is similar to that a sphere, i think)

for the plane

$$V = q \left(\frac{1}{\sqrt{x^2 + y^2 + (z-z_{0}}^2}} - \frac{1}{\sqrt{x^2 + y^2 + (z+z_{0})^2}}\right)$$

for the hemisphere ... satisifes all those requirement of a sphere for y > 0

$$V = q \left(\frac{1}{\sqrt{r^2 + z_{0}^2 - 2rz_{0}\cos\theta}} - \frac{a/z_{0}}{\sqrt{r^2 + (a^2/z_{0})^2 - 2r(a^2/z_{0})\cos\theta}}$$

where d
i think i t would be better to convert to cartesian coords
then the solution for the hemisphere is
$$V = q \left(\frac{1}{\sqrt{x^2 + y^2 + (z-z_{0})^2}} - \frac{(a^2/z_{0})}{\sqrt{x^2 + y^2 + (z-(a^2/z_{0})^2}}\right)$$

the potential is then
$$V = q\left( \frac{1}{\sqrt{x^2 + y^2 + (z-z_{0})^2}} - \frac{1}{\sqrt{x^2 + y^2 + (z+z_{0})^2}} + \frac{1}{\sqrt{x^2 + y^2 + (z-z_{0})^2}} - \frac{(a^2/z_{0})}{\sqrt{x^2 + y^2 + (z-(a^2/z_{0})^2}} \right)$$

this will not satisfy the boundary condtion when y>=a and y<= -a entirely as the 2nd and 3rd terms will not disappear.

need another charge to cancel those terms which remain
$$V(0,y,0) = q\left( \frac{1}{\sqrt{y^2+z_{0}^2}} - \frac{q(a^2/z_{0})}{\sqrt{y^2+(a^2/z_{0})^2}} \right)$$

would i add some charge q' located ebtween the two image charges to cancel out the remaining terms??

5. Jan 16, 2007

### mjsd

As u can see, you need more and more charges, that's why I suggested a series of images. I must admit I haven't tried solving this in full, but if i were you that's the step I will try:
continue adding in images to "fix up" any problems, check and see whether the magnitude of the newly included charges are progressively getting smaller, if so i suspect that the series would converge as your charges should be alternating in sign, if not, they may be conditionally convergent (much harder to prove sometimes) good luck

6. Jan 16, 2007

### Meir Achuz

1. To make the plane grounded, put -q at -z_0.
2. To make the sphere grounded put image charges -qa/z_0 and +qaz_0
at +a^2/z_0 and -a^2/z_0.

7. Jan 16, 2007

### mjsd

mmm... so the series terminates after 3 terms? Interesting.

8. Jan 17, 2007

### chanvincent

Since you know the answer for the individual, why don't apply the super-position principle on them to get the answer for the "mixed" case?