# Using the method of Mirrored Charges to find the charge force due to grounded sphere on 2 charges

1. Nov 10, 2014

### Yosty22

1. The problem statement, all variables and given/known data
Two point charges of equal value are placed a distance 2b from one another. In the middle, there is a grounded conducting sphere. What is the minimum radius of the sphere required to cancel out the repelling force between the two charges? (Solve using Image Charge Method)

2. Relevant equations

3. The attempt at a solution

I attached a file I used to create the scenario in. During one of the lectures, we proved what I wrote on the right-hand side of the file to show that in order to keep the conducting sphere at a potential of 0, you must put a negative charge at a distance R^2/r from the center of the sphere. Since this is spherically symmetric and you are adding another positive charge outside of the sphere (the charge labeled q2), could you just add another mirrored charge a distance R^2/r from the center of the sphere, except this time, in the negative direction? That is, add a charge of -q at a position R^2/r to the left of the center.

Therefore, your total setup for the mirrored charges would be one charge of q = -q located a distance R^2/r from the center of the sphere in the positive x-direction and another charge q = -q' located a distance R^2/r from the center of the sphere in the negative x-direction?

#### Attached Files:

• ###### Picture.docx
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2. Nov 11, 2014

### BvU

I agree with the approach: the potentials from the two configurations can be added up.
I miss the relevant equation, but never mind.
Symmetry isn't spherical but rotational. Plus there is mirror symmetry. Never mind that either.
Given is that the charges are equal, makes is somewhat more manageable.
So q1 = q2 = q.
They are at distance 2b which became 2d. Fine too.

Mirror charge value is also known; it's not -q.

All in all you have four charges lined up at -d, -R2/d, R2/d, and d.
Your job is to say something about R in the case the field at -d and +d has to be zero.

That's where I end up with a hefty equation for R/d that I can only solve numerically. And why the exercise says minimum R is weird to me: there is only one R that causes this equilibrium.

Please check and correct me :)