# Method of Images: Grounded Conducting Sphere

1. Feb 15, 2013

### darwin59

1. The problem statement, all variables and given/known data
A point charge +q is situated a distance a from the center of a grounded conducting sphere of radius R. Find the induced surface charge on the sphere as a function of θ. Integrate this to get the total induced charge.

2. Relevant equations
σ=-εo*∂V/∂r
Q=∫σ da

3. The attempt at a solution
I've already figured out σ to be (-q/4∏)(a2/R-R)(R2+a2-2aRcosθ)-3/2. I also know that Q needs to equal -q. But I'm having trouble completing the integral to get to just -q. The way I've tried it is ∫$^{2∏}_{0}$∫$^{∏}_{0}$ σ R2sinθ dθd$\Phi$; but I'm not getting it to reduce to just -q. I'm confident in my answer for σ, so my problem must be in my integral, but I can't figure out what it is.

Last edited: Feb 15, 2013
2. Feb 15, 2013

### darwin59

Well I worked it down further, and I got it to be +q, not -q, so I must be missing an extra negative somewhere. Here are my steps:

∫d$\Phi$ from 0 to 2∏ is just 2∏
I used u substitution, with u=R2+a2-2aRcosθ, and du=2aRsinθdθ.
So now I have Q=(-qR/2)(1/2aR)(a2-R2)∫u-3/2du.
Q=(-q/4a)(-2)(a2-R2)u-1/2=(-q/2a)(R2-a2)(R2+a2-2aRcosθ) from 0 to ∏.

Q=(-q/2a)(R2-a2)[1/√(R2+a2+2aR)-1√(R2+a2-2aR)]

Q=(-q/2a)(R2-a2)[1/(R+a)-1/(R-a)]

Q=(-q/2a)(R2-a2)[(R-a-R-a)/(R2-a2)]

Q=(-q/2a)(R-a-R-a)=(-q/2a)(-2a)=q

Did I miss a negative?

3. Feb 15, 2013

### TSny

Check the step where you evaluated √(R2+a2-2aR). Recall √(x2) = |x|

4. Feb 15, 2013

### TSny

Are you sure of this? For the case where the charge q is very far from the sphere, would you expect the charge q to be able to induce a total charge of -q on the sphere?

5. Feb 15, 2013

### darwin59

We're assuming it's close enough to induce a total charge of -q on the sphere.

6. Feb 15, 2013

### darwin59

I'm not sure how that would help. I could rearrange that as (-q/2a)(R2-a2)[1/(a+R)-1/(a-R)] to take care of the absolute value, but then I'll end up with (-q/2a)(R2-a2)/[(a-R-a-R)/(a2-R2), and I'm stuck with Q=-qR/a.

Actually, come to think of it...that's what the charge of the image is supposed to be, according to my book. Now I'm confused because some homework help I've found says the total charge should be -q, but it doesn't explain how it got there. -q is what I would expect, but at the same time, it's hard to imagine the distance from the sphere not influencing the induced charge on it, so maybe it is -qR/a. Any thoughts?

7. Feb 15, 2013

### TSny

Apply Gauss' law to any closed surface that encloses the sphere but not the charge q; for example, you could choose a spherical surface concentric with the conducting sphere but just a bit larger.