# Homework Help: Charge accumulation using radiation

1. Feb 4, 2014

### mooreppj

Hello! I've been working on this physics problem for a physics II course and I'm stumped. Maybe someone can assist?

1. The problem statement, all variables and given/known data

A small insulated copper sphere of radius 0.001m is illuminated with radiation with wavelength 300nm. What is the maximum charge accumulation on this sphere?

2. Relevant equations

The only equation I can do unit analysis on and get a coherent answer is E = hc/λ
I'm not sure which equation to use next, however I think that:

E = kq/r^2

is an important equation for this problem.

Finally, the surface area of a sphere:

A = 4$\pi$r^2

3. The attempt at a solution

Initially, I calculated the energy associated with the radiation:

E = hc/λ = hc/(3e-7) = 6.62e-19 J

Then, I divided the sphere's surface area (A = 4$\pi$r^2 = 1.26e-5 m^2) to get a flux (also apparently called irradiance) of light over the sphere's surface.

Flux = 5.27e-14 J/m^2

Now, I'm not sure how to proceed. I'm having difficulty conceptualizing how irradiance influences charge on an insulated object. I thought that in order for charge to accumulate on an insulated object, an electric field must be present. My understanding is that light is electrically neutral and by itself does not create electric fields. Could the copper substrate be significant in creating a charge through being ionized? I feel like that point is beyond the scope of the question but I'm completely lost on how to proceed.

Thanks for any help anyone can offer!

2. Feb 4, 2014

### BvU

Hello mooreppj, and welcome to PF. Lovely problem you are faced with. If your rendering is complete, some assumptions might be necessary. You judge if that's nonsense or not.
First, if you illuminate with radiation and they are asking for a maximum charge: what kind of radiation ?
I'm glad to see that (after some skirmishing with equations) you pose the question as well. Good. My instinct (proven to be good sometimes, worthless at other times) tells me you've been introduced to some kinds of radiation in the preceding chapter already. Met your new friends α, β, and γ. Originally I naively hoped that the middle one would do the trick. No way. Wavelengths 300 nm are for photons only. No idea how they can charge a 1 mm copper pellet?

So it's down to you: what's the context? What's the preceding chapter about ? Are we doing electrostatics here (like: kinetic energy has to overcome potential difference). Something else ? Photoelectric effect, for instance ?

I'm intrigued, puzzled and interested.

Can only be photoelectric effect. Photons knock out electrons until potential > Ekin max.

Last edited: Feb 4, 2014
3. Feb 4, 2014

### mooreppj

Hey BvU, thanks for responding so quickly!

I'm certainly willing to use some assumptions if they're reasonable. That's the question as written unfortunately, so I think SOME assumptions need to be made. I don't think there is anything particularly novel about the radiation used. If I've made an assumption, it is that the radiation uniformly irradiates the sphere from all directions. I applied this assumption when I divided by the surface area of the sphere to get the flux of photons per squared meter.

Other than that though, I can't think of any other assumptions that are reasonable to solve the question.

We've recently done several chapters on particle physics, nuclear physics, and introductory quantum physics. We've only used electrostatics insomuch to describe charged particle phenomenon such as kinetics with electrons, protons, neutrons, and the particle zoo. I don't think there are any waveforms from nuclear particles though because the radiation is in the ultraviolet spectrum.

When he asks for the maximum charge accumulation on this sphere, I wish I could be more specific. We've done material surrounding blackbody radiation, the photoelectric effect, compton, de broglie, etc etc. I'm sure you get the gist.

As the question is asked though, when I see "accumulation" I think energy balance:

dQ/dt = Charge in - Charge out,

Where because the sphere is insulated, would not release any charge to the environment and

dQ/dt = Charge in

But the fact that nothing is mentioned in the question about time kind of stops that thinking in its tracks. I also can't figure out how, as the question is written, 300nm radiation is charging anything. I feel like if I flashed a UV light all around a copper pellet in a dark room, it wouldn't ever get a charge. Could this be a trick question?

4. Feb 4, 2014

### mooreppj

I should also note that I don't think the photoelectric effect applies for a copper sphere at this particular wavelength. Using the following equation for the photoelectric effect:

Kmax = hf - θ

Where theta is the work function. In this case, it is copper which my text ( Serway & Jewett, 9th ed.) says is 4.70 eV. Doing all the math to solve for frequency:

f = c/λ = 9.993e14 hz

And solving Kmax,

Kmax = (6.626e-34) x (9.993e14) - [(4.70 eV) * 1.602e-19 J/eV]
Kmax = -9.089e-20

The Kmax being negative tells me that the wavelength of the radiation is too long to remove electrons from the surface of the copper sphere. In fact, if I solve for the cutoff wavelength:

λ(cutoff) = hc/θ
λ(cutoff) = 264 nm

I think I understood what your point was though; that if the electrons could be ejected from the surface the sphere, the charge on the sphere would be equal in magnitude to the charge of the electrons ejected. However I don't think that it works in this case for the reasons I worked out above.

5. Feb 4, 2014

### TSny

At first I thought this might be a nice problem involving the photoelectric effect with photons knocking electrons out of the copper. After some calculation, I have some doubts. Were you given a work function value for copper?

[EDIT: oops, I see you have already addressed this!]

6. Feb 5, 2014

### BvU

Dear mooreppj: Congrats! You have rummaged through the information given, done some checks that convinced you this can't work (λcutoff). You're not omniscient, so you can't be completely sure and the best statement you can make is exactly what you write down.

I found a comparable exercise (problem 28.13) but figured you were able to get through this on your own (I'm not omniscient either, but definitely lazy -- basically i missed the λ difference, so I didn't bother to actually do the calculation -- but what the heck).

So your question as written is either a trick question, or some oversight by a lazy teacher or exercise writer. I have strong views in both cases:
Trick questions are didactically treacherous except in direct communication, where body language can get things right. On paper, you just feel, well, tricked.
Lazy physicists aren't necessarily bad physicists (I surely hope ;-) because they tend to think a little longer before doing unnecessary work.

In this case you have risen well above the level of a simple exercise. Enough flattery by now, get back to work and tackle the next one!

Oh, and: if I turn out to be completely wrong I would more than appreciate a posting of what was supposed to come out!