# Induced charge distribution in spherical simmetry

I've been trying to understand the following situation for a while now. It seems something so simple and easy and yet I can't find answers. Suppose we have a configuration like the one shown in the picture.

We have a charge q surrounded by a spherical conductor of radios A and B.

I understand how to find both the E-field and the potential for every region, but I don't understand what's happening with the charge inside the conductor. Of course inside a conductor E=0. The charge will then distribute itself on both surfaces to achieve this; thus, on the smaller surface I'll have an induced -q charge, and on the outer surface, I'll have an induced q charge.

But what about in the proximity of both surfaces inside the conductor? Would they concentrate on both surfaces and then simply "disappear"?(Kind of like a discontinuity, as the orange function shows) or is it a more gradual and continuous process, where most of the charge resides on these surfaces and then fastly starts to decay to 0? (as the green function on the same region is trying to represent)?

If I then connected the conductor to a battery of voltage V=V0, would the potential affect the charge distribution in any way? I feel like it should, since q= V * r * (4π) but then again, I still need to cancel out the q charge in the center, so it should remain all the same.

Thank you to anyone wanting to help.

If you assume that the charge density is 2-dimensional (as is often done in basic EM courses), then yes, the integrated charge at radius $$r$$ has discontinuities at $$r=A$$ and $$r=B$$. Working our way up the complexity scale, the charge distribution is actually 3 dimensional and will change a bit more smoothly "in reality," as real metals are not perfect conductors.

Funnily enough, if you continue working your way up the complexity scale, you'll have to acknowledge that we don't actually expect the integrated charge function to be continuous, since charges are discrete and quantized! And further up the complexity scale, we note that quantum mechanics dictates that the positions of charges aren't even well defined, so the concept of a well defined $$Q(r)$$ function doesn't strictly exist.

With all of that said, the most important thing to gather is that everything you learn in EM is just a model of reality. You can think of the induced charge distribution as either 2D or 3D as long as it leads you to the correct results. The reality is so much more complicated that we can't even use first principles to calculate it

EDIT: If you connect the outer sphere to a battery, then yes, the charge distribution will change. You should look up the "method of images for a sphere at a constant potential" for the (very interesting) mathematics behind this.

• Hello @Dr Momo. Thank you for your answer! Just a tiny question, is the method of images the only way to find the charge distribution when the outer sphere is connected? I've been searching for this method. I feel I get the idea behind it, but I've only found problems where the q charge is outside the conductor and I don't see the relation with my problem. Commented Mar 10, 2022 at 2:39

Just wanted to follow up with some insights. I believe that I was mistaken -- your last question is best answered using techniques outside of the method of images. This is, to the best of my knowledge, correct, but a subject-matter expert should probably verify this. I want to know if I'm right, too, so consider reposting the last question only.

Consider a hollow spherical conductor of inner radius $$R_A$$ and outer radius $$R_B$$. A charge $$q$$ is placed in the center, and a conductor is given a potential $$V$$. There are three regions to consider: inside the conductor, the conductor itself, and outside the conductor. Our goal is to find the potential in the three regions.

The solution inside the conductor is trivial: $$V_{II}=V_0$$.

The potential inside of the spherical cavity is calculated by the method of superposition. The potential due to the charge itself is $$\frac{q}{4\pi\epsilon_0}\frac{1}{r}$$ while the potential due to the two surface charge densities ($$\sigma_A$$ and $$\sigma_B$$) is a constant. Thus:

$$V_{I}=\alpha+\frac{q}{4\pi\epsilon_0}\frac{1}{r}$$

The potential is continuous at $$r=R_a$$ (it equals $$V_0$$ there), and we can use this to eliminate $$\alpha$$. Thus:

$$V_{I}=V_0+\frac{q}{4\pi\epsilon_0}\left(\frac{1}{r}-\frac{1}{R_A}\right)$$

This equation seems right to me, since the boundary condition holds and it seems to satisfy Poisson's equation:

$$\nabla^2V=\frac{q}{4\pi\epsilon_0}\nabla^2 \frac{1}{r}=\frac{q}{4\pi\epsilon_0}\left(-4\pi\delta^3(\mathbf{r})\right)=-\frac{q }{\epsilon_0}\delta^3(\mathbf{r})=-\frac{\rho}{\epsilon_0}$$

The charge density at $$r=R_A$$ can be derived from the following equation in Griffiths Introduction to Electrodynamics (Third edition, Section 2.3.5):

$$\nabla V_{above}(R_A)-\nabla V_{below}(R_A)=-\frac{\sigma_A}{\epsilon_0}\mathbf{\hat n}$$

I'll spare you the details, but I get

$$\sigma_A=-\frac{q}{4\pi R_{A}^2}$$

As it turns out, the charge distribution on the inner surface does not depend on the applied potential. Additionally, you can see that the total induced charge on the inner surface is $$-q$$, just enough to completely neutralize the inner charge.

The potential outside of the sphere can be derived by considering the following general solution given by Griffiths (Third Edition, Section 3.3.2, Example 3.7):

$$V=\sum_{l=0}^\infty \left(A_l r^l + \frac{B_l}{r^{l+1}} \right)P_l(\cos{\theta})$$

As before, I'll spare you the details, but it turns out that

$$V_{III}=\frac{R_B V_0}{r}$$

The surface charge density on the outer surface can be derived as before. I get

$$\sigma_B=\frac{\epsilon_0 V_0}{R_B}$$

This seems to make sense -- if the sphere is grounded, then no charge will be induced. And as $$V_0$$ increases, the induced charge increases.

I'm sure that others can give a 2-sentence argument for why the above had to be true intuitively, but there it is, a priori.

• From the bottom of my heart, thanks. While reading I was making the calculations myself, thankfully I reach the same results. I'm extremely happy, It's been around 4 or 5 months since I was first introduced to this type of problem and never understood how a conductor with a given potential would affect the situation in comparison to a disconnected one. Commented Mar 13, 2022 at 2:07
• Not a problem at all! I had fun putting this together. Commented Mar 13, 2022 at 20:05