# Detecting the charge distribution within a Faraday cage

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1. May 1, 2015

### Happiness

Imagine an uncharged solid spherical conductor. Inside this spherical conductor, there is a cavity of a weird shape carved out of it. And somewhere inside this cavity, there is a charge +q (or rather, a charge distribution of total charge +q).

The charge +q induces an opposite charge -q on the wall of the cavity of the conductor, which distributes itself in a such way that its electric field cancels out that of +q for all points exterior to the cavity. Since the conductor carries no charge, this leaves +q to distribute itself uniformly over the surface of the sphere.

How can we detect or figure out the charge distribution in the cavity? Since the charge distribution on the outer surface is uniform, measuring the electric field won't work.

Ultrasound and x-ray have been suggested. But I suppose they are meant for detecting mass distribution because I don't see how ultrasound and x-ray can detect a charge distribution (within a cavity of a spherical conductor). I suppose mass distribution and charge distribution are not necessarily related. Ultrasound and x-ray are probably good for detecting the size and the location of the cavity, but they may not be good for detecting the charge distribution in it.

I thought about perturbing or varying the ambient electric field in which the sphere is placed, thinking that a sphere with a cavity would react differently from another one that has a different charge distribution in its cavity. By reacting differently, I mean the surface charges on the spheres would move differently. And hence, by observing their motion or re-distribution during a perturbation of the ambient electric field, we can, in principle, deduce the contents or structures underneath.

But on second thought, I think the spheres won't react differently since the surface charge of the cavity cancels the effect of the charge in it completely.

Before we try to figure out a method of detection, it makes more sense if we first answer this question: Does the requirement of a uniform charge distribution on the outer surface constitute a no-go theorem? In other words, does physics say it's impossible to detect the charge distribution within the cavity (unless you cut open the sphere)?

Lastly, for emphasis:

Is it a no-go theorem? Is it a no-go theorem? Is it a no-go theorem?

2. May 1, 2015

### rumborak

The reason x-rays and the like are suggested is, Faraday cages have an upper frequency limit, above which they no longer function (because it takes the charges too long to get into equilibrium). So, once you get through the cage, you could probably detect what's inside.

3. May 1, 2015

### Happiness

Does it mean that if we vary the ambient electric field at a frequency above the upper frequency limit, we can detect the charge distribution inside? But the surface charge on the surface of the cavity is already in equilibrium with the charge distribution in it, so I suppose the ambient electric field will interact with and affect only the surface charge on the outer surface as though the cavity and the charge distribution in it are not present. Or is this false? When the frequency of the varying ambient electric field is above the limit, the electric field will penetrate the conductor and distort the surface charge distribution of the cavity. As a result, the cancellation of the effect of the charge distribution in the cavity is no longer complete, and then, by a measurement of the electric field, we can determine the charge distribution. Is this correct?

Could you explain more about the x-ray method? Is it the greater the charge density, the greater the absorption or deflection of x-ray?

Last edited: May 1, 2015
4. May 1, 2015

### rumborak

I'm not particularly claiming a practical method for detecting charges inside a Faraday cage, I'm just saying that above the frequency limit, the signal attenuation is no longer complete. Which in theory should make it possible to detect what's inside.
Interestingly, there's also a lower limit. The cage also relies on something called the "skin effect", which has a certain depth. If the wavelength of the incoming signal exceeds that of the thickness of the wall, it can also no longer do a complete job at attenuation.

I wonder though, what's your obsession with detecting charges in a Faraday cage?

5. May 1, 2015

### Happiness

It seems that I am allowed to answer that question. Apparently, any discussion that is not technical will get the thread locked.

6. May 1, 2015

### Staff: Mentor

This result may be offensive to your intuition, but the universe doesn't seem to care about that. Physics is about the way universe is not the way it ought to be, and Physics Forums is run accordingly. Thus, we will cheerfully discuss the properties of the solutions of $\nabla^2\phi=\rho$, which is the relevant equation here, but we won't allow arguments that the universe is "wrong" for obeying that equation.