timo

Most photons have spin 1, btw.

Slightly related, potentially interesting in this scope, not checked by me for actual physics content or credibility: http://www.compadre.org/informal/index.cfm?Issue=2 , "Flames in Space".

Wouldn't that depend on the value of [math]\Delta[/math], too?

I would consider the 4-periodicity of exp(ix) under derivatives (i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, i^5=i^1=i) quite an important characteristic for reproducing the sine and the cosine which also have this 4-periodicity under derivatives.

One of the bad aspects of socialism is that you need to build big anti-fascist protection walls to protect your people from the evil capitalists on the other side of the border.

Shouldn't that argument work for electrons, too? Particularly in the sense of the conclusion that electrons were their own anti-particles?

I don't really see why you did not follow my advice. Even if I had the Windows-only file conio.h installed on my system I really doubt your code would even compile; lines like "int date();" in a source block just seem wrong to me. Even if your code did compile you should at least see warnings like the one that tells you you should include <iostream> instead of <iostream.h>. And I still think you should comment each step your program is supposed to do. While it seems clear to me that the mentioned "int date();" should not compile you probably had a reason to add that line but it is absolutely unclear why you thought it should be there and what you wanted to achieve with this line.

- I do not understand your calculation for the potential V®: The (3 pi epsilon)^-1 probably is just a typo, so no problem with that. But I do not understand where the kr'/r' part comes from. What is the r' in the denominator? - You are missing integration ranges in your final calculation. This causes your result to be expressed in terms of an arbitrary variable r rather than the given radius of the sphere R.

I am not sure if you really understood what the question asks: If you have two charges at some distance then there is some energy gained by seperating the to an asymptotically infinite distance (a negative value if their charges are opposite). The same holds true for three charges, four charges, ... . That is this pretty much the definition of potential energy. The same is supposed to hold true for infinitesimal charges [math]\rho \, dV[/math]. Calculating the potential energy for a set of charges [math]\rho \, dV[/math] is what the question asks for. For two charges, the potential energy is something like [math]C \frac{q_1 q_2}{\| \vec x_1 - \vec x_2 \|}[/math], with C being some suitable constant. For n charges the expression becomes [math] \sum_{i<j} C \frac{q_i q_j}{\| \vec x_i - \vec x_j \|} = \frac{1}{2} \sum_{i \neq j} C \frac{q_i q_j}{\| \vec x_i - \vec x_j \|} [/math]. In the limit of infinitesimal charges this becomes [math] \frac{1}{2} \int_V dV_1 \rho(\vec x_1) \underbrace{\int_V dV_2 C \frac{\rho(\vec x_2)}{\|\vec x_1 - \vec x_2 \|}}_{=: V(\vec x_1)} = \frac{1}{2} \int_V dV_1 \rho(\vec x_1) V(\vec x_1)[/math]; the expression you are asked to calculate. So your first step, finding an analytic expression for V, is fine (I did not check the calculation; from my experience most students are better at rearranging equations and integrating stuff than me, anyways). However, since the charge density is zero outside the sphere, I do not understand why you bother with this volume at all - it will contribute a plain zero. I think your problem might lie in understanding what you are supposed to calculate and why so - hence my little handwaving explanation above.

They do not obey the energy-momentum relation that a free particle of the respective field would have.

- You can use the [ code] <code here> [ /code] brackets to make the code more readable: #include <iostream> // for text input and output using namespace std; // std::cout -> cout int main() { // Ask user to guess a number. cout<<"Enter a guess for the number of errors in your code: "; int errors; cin>>errors; // React on ridiculous guesses. if (errors==0) cout<<"You wish!"<<endl; } - You should comment your code; you did not even say what it is supposed to do (my example did not either - because it is silly). - Start with the snippet I gave you and replace parts of my code with yours. Your code is so erroneous I don't think it is worth keeping. - Utter-newbie tip: Compile early, compile often. If two new lines make the previously-compiling code fail, it must be related to those two lines.

As said: The [math]\hat z[/math] in the integrand is wrong. The rest seems ok.

To be honest: Yes, it is solvable in the sense that there is an analytic expression for it - I did it with Mathematica . Math handbooks might have it as a standard integral in their tables but I couldn't check that as my math handbook lies in my office. The charge density never is a vector, regardless whether it is charge per volume, per area or per 1D-volume.

That is definitely better. A few comments/corrections, though: - [math] \lambda \, dz[/math] is not a charge density. If you consider dz being a small line segment (the typical view physicists take when dealing with differentials) then it is a small charge. Technically, the density would be [math] \rho(x,y,z) = \lambda \delta (x) \delta (y) \Theta(-z) \Theta (L+z)[/math]. If you don't know the delta-distribution and the theta-function then just ignore this comment (it is just a complicated way of saying "zero everywhere except on the line segment from the origin to (0,0,-L) where it equals lambda"). The point is merely about units, anyways. - Where does the [math] \hat z [/math] come from under the integral? Both, r' and r are mere real values. They should not suddenly become a vector.

- the TeX code for vectors is \vec, e.g. \vec a -> [math]\vec a[/math]. - Using your expressions for [math]\vec x[/math] and [math]\vec r[/math] then [math]\| \vec r - \vec x \| = \sqrt{ (0-a)^2 + (0-b)^2 + (-z -c)^2} = \sqrt{ a^2 + b^2 + (z + c)^2}[/math]. Not the term you wrote. - I'd recommend being consistent with signs and use [math] \vec r = +z \hat z[/math] (this will alter the expression for [math]\| \vec r - \vec x \|[/math]!). Your integration range (from -L to 0) indicates that you accidentially switched to it somewhere along the way, anyways. - The [math]\frac{\rho ®}{r} [/math] integrand still looks strange to me, to say the least. What is r supposed to stand for (in terms of physics)? Is it an R³ vector or a real value? I think your confusion might stem from not having completely understood the expression for the electrostatic potential.

[math] V(\vec r) = C \int_{\vec x \in V} dV \frac{\rho(\vec x)}{\| \vec r - \vec x \|} [/math]. Your two rs in the integrand refer to different locations (the position of the generating charge and the distance from the position). The equation but not be sensible, otherwise. Of course this relatively general expression simplifies to a 1D integral in your case; I did not rewrite it that way because you might want to figure it out yourself (assuming the notation in your integral was the problem). EDIT: Oh, sorry: Your expression can make sense if you shift your coordinate system appropriately, i.e. if you look at V(\vec 0) for a charge distribution in a line from (-x,-y,-z) to (-x,-y-L,-z) (not exactly sure about the coordinates; due to my currently crappy INet connection I cannot see the original problem).

The equation q=Fd/V is correct in this case. The reasoning is a bit thin (equations not justified or explained) but I suppose it is sufficient for school level - and well, as I said, they are correct. For the actual result: You did not provide any units for your charge and the voltage. Assuming the voltage is 310 V then with your equation I'd get a charge of approximately q = Fd/V = 8.4*10^(-7) N * 0.2 m / 310 V = 1.68 * 10^(-7) J / 310 V ~= 1.68 * 10^(-9) C / 3 = 5.6*10^(-10) C. Except for the exponent that's pretty close to your answer. Never omit the units!

It's not exactly the idea of the homework section to walk you through the question. You should at least state where your problems with the question lie, what ideas you have, what ideas for solving the problem you have (and why you think they will not work) and what equations/symmetries/properties/... you think might be applicable/relevant here. As a starter: Do you know how potential, potential energy, electric field and force on a charge in an electric field relate to each other? Do you know what the electric field for "the region between oppositely charged parallel plates" looks like?

Did you mean kg m²/s² or does it already answer your question to tell you that there's a factor m (for mass), too? The units of relativistic energies are just the same as for non-relativisitc ones.

You'll have to look at the forces (the acceleration will follow from them via a=F/m later) for this problem.

So? What's your ideas on how to tackle the problem? Can you calculate theta or at least have an idea how it could be done or what it would depend on? If you knew theta could you calculate the acceleration? Could you calculate the acceleration in the absence of friction? ...?

Why do you think it could not be vacuum, i.e. no other molecules/atoms? If you skip all unnecessary complication then "vacuum", in the sense of "nothing", is a decent answer to the question. Note however, that the term "vacuum" is usually used on larger length-scales and also for volumes that can have a few (but relatively little) molecules in them.

That would mean that the threshold lies somewhere in the range of the user title "atom" or below, I would think.

Perhaps an example of what you mean would help? If you are talking about reciting stuff (derivations, definitions, ...) that were given in the course/lecture/class/...then a standard way of preparing is making cards with the respective keywords, mix the cards, draw them at random, write down the derivation/definition/... for the keyword and, in case you cannot, look it up. This procedure is then repeated until you can write down them all without looking something up (potentially by removing cards which become too boring in between). Not sure if that is what you are talking about, so ... see my first sentence (note also that terms like "objective exams" and "subjective exams" might have little to no meaning for people not in the same country/area/school/education level as you).

Yes. That's the main reason I don't like these kind of pictures "explaining" gravitational effect as a result of spacetime curvature. By that I mean pictures like this one: "Understanding" this picture -in my opinion- heavily relies on the (knee-jerk) assumption that things have the tendency to fall down. However, from the mathematical standpoint it is only the geometry of the plane that determines motion on the plane. Hence, the picture is equally valid if you turn it upside down. The attached picture is the same one as above, just rotated by 180°. In my opinion, it is absolutely not clear from the visualization why the smaller peak (smaller object) should circle upwards to the larger peak (larger object) now. I think something is fishy with this rubber-plane visualization. I don't really know what, though. As for the implication on science that these visualizations have: Practically none. When you do calculations you do not rely on terms like "dents in spacetime" but rather just calculate some properties of spacetime, apply some rather general criterion (path of an object is such that some quantity integrated over the path is minimal) that depends on these properties and get a result that coincides with the observed phenomena. No referal to gravity, there. Only the visualizations made seem to rely on our everyday-knowledge of Newtonian Gravity.