1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Charged Particles

  1. Dec 31, 2008 #1
    I'm hating Electromag as of late. To the extent I can't even understand these 'simplest' of questions. I don't want anyone to be doing my work for me as nothing is essentially accomplished that way. If someone could head me in the right direction with these questions, I'd be most grateful.


    My answers (using some rudimentary calculation) came to:

    i) c - [-30v]

    Further away than y and therefore more negative but not by much thus -30v, this does not provide enough substantial evidence of mathematical precision, rather just a logical guess.

    ii) a - [surface charge density is constant]

    iii) b - [zero potential]

    iv) d - [2.3 x 10^29]

    v) c - [other forces]

    Kind Regards,
    Last edited: Jan 1, 2009
  2. jcsd
  3. Jan 1, 2009 #2
    Bump - Can anyone help me on this?

  4. Jan 1, 2009 #3
    Hello Billy246 :smile:

    Well i) is simple if you just use V = kq/r, or 1/(4pi epsilon) * q/r, whichever notation you like. Now, you got ii) right, but can you prove your answer? (Hint: use Gauss's Law) Try looking at iii) again... does it HAVE to be at 0 potential?

    I'll take a look at the rest a bit later... I'm in a bit of a hurry now, so until then, good luck! :smile:
  5. Jan 1, 2009 #4
    Thanks with this, your assistance is gratefully received. I don't want to appear pushy on these forums as it is other people's time being used.

    Basically used your idea on
    Part i
    Potential = kq/r = [(8.99x10^9) x 30nC]/3

    Then for charge point 2?

    [(8.99x10^9) x -20nC]/root 13

    add the two together to get ~40V

    thus the answer is a)

    Part ii
    The answer I gave was only due to the logicality of it. I cannot provide an adequate reasoning for my answer.

    Part iii
    I was under the false impression that it must be at zero potential.
    It must have something (this therefore rules out 'neither'), is the charged hollow sphere at a constant potential due to the constant surface charge density?

    Kind Regards,

  6. Jan 1, 2009 #5
    Okay, so it looks like you got the first question right. Good job! :smile: For the next part, try to use Gauss's Law. If you haven't heard of this, then never mind... just know that whenever you have a conducting surface, the charge distributes itself over the surface such that *the net electric field within the conductor is ZERO.* This is absolutely essential, because we can't have currents present within a conductor for absolutely no reason.

    Now, for the next part with potentials over a conducting surface, note that potentials are always relative to another. In other words, we only care about potential differences. So, stating that something is AT 0 V potential means absolutely nothing; this ONLY means something when you say that the potential at infinity (the potential really really really far away) is 0. Now, the answer is actually that the potential over the conducting surface is a constant. Try to see what happens if the potential across a conducting surface is NOT constant: if two areas have different potentials, then electrons will flow between them, causing a current! Eventually, everything becomes balanced, and no electrons flow, implying no potential differences across the surface of the conductor.

    Does this make sense? :smile: If not, please say so, and I will try to help you better. Good luck, and happy new year!
  7. Jan 1, 2009 #6
    Yes thank you, you have provided me with enough information to understand the first three question parts.

    I have a problem visualising it all, my physics lies in space not charges!

    If you could point me in the right direction for the final two parts I would be most thankful.

    Thanks for your time and kindness.

  8. Jan 2, 2009 #7
    iii) b - [zero potential]

    iv) d - [2.3 x 10^29]

    v) c - [other forces]

    Kind Regards,

    [Hi Billy I guess your at the uni of leicester as I have the exact same questions to do. I've got these answers so far

    i) a
    ii) a
    iii) a
    iv) c (1.8x10^29) - I got a different answer for this than what you got. How did you get your answer?
    v) i'm not sure about this how did you come to your conclusion for this?

    Thanks Ben]
  9. Jan 2, 2009 #8
    Yeah I am, finding this paper tricky though.

    As for parts iv and v, i'm fairly sure iv is wrong due to my highly crackpot calculations. I don't even think I have my 'working' anymore. I wouldn't submit the answer if I were unsure.
    On v, I don't understand it but I thought other forces were in play, I don't know if it's ohmic or not :S


    Looking at it now, the material must be ohmic as this proportionality comes from the fact that an applied electric field superimposes a small drift velocity on the free electrons in a metal.
    Last edited: Jan 2, 2009
  10. Jan 2, 2009 #9
    I see what you mean. Tipler isn't being overly helpful but it kind of suggests the material being ohmic like you have suggested but what im not sure about is whether the answer could be (d) or (e)?? How have you found the ret of this paper and the others?
  11. Jan 2, 2009 #10
    Read my pm, I can provide you with some help.

    With part iv, i'm inclined to ask you for your working. How did you get to that value. And part v is a toss up between a,d,e :/

    Maybe someone here can assist us with this.


    Basically with iv)

    (2.7 x 10^3)/27 = 100

    100 x (2.7 x 10^3) x 6.02 x 10^23 = 1.625 x 10^29

    ~1.6 > b.

    However not sure how to factor in the fact that there are 3 carriers per atom
    Last edited: Jan 2, 2009
  12. Jan 2, 2009 #11
    well with the information given

    M = 27.0gm/mol = 0.027kg/mol
    ro = 2.7 x10^3 kg/m3
    N_a = 6.02 x10^23 (atoms)/mol (it doesnt tell you the atoms part but I think that might jus be seen as common sense if you did chem at a level)

    first I calculated the mass of the Al atoms by:

    mass = N_a/M = 2.23 x10^25 atoms/kg

    then it says in the question there is 3 e- per atom so i multiplied the mass by three (called it mass_e)

    mass_e = 6.69 x10^25 carriers/kg

    Lastly the number of carriers in the wire (n) in that density =

    n = ro x mass_e = 1.8 x10^29 carriers

    I checked the units and they work out right.

    Hope that helps.
  13. Jan 2, 2009 #12
    I just read part V, and I think the answer is (C). Here is my reasoning (I may be overthinking the problem, but this is true): the problem doesn't state that the current is *immediately* steady, but is *quickly* steady. From this, we can observe that the voltage must increase from 0 to V at a very quick rate. Now, when we have an changing voltage, we have a changing electric field, which causes a changing magnetic field to exist, which causes another changing electric field to exist, which causes another changing magnetic field to exist, etc. Now all these fields (not only electric, but magnetic since the charged carriers always have non-zero velocity... use the Lorentz force) act on the charged carriers, so we have that there must be other forces exerted on the charge carriers besides that due to E.

    The material actually does not have to be ohmic (meaning that is has resistance); if we have a thin wire which has magnetic fields on the edges such that electrons are always repelled and whose interior is a vacuum (minus our charge carriers within), for example, establishing a voltage difference still proves us with a steady current, and this wire is clearly not ohmic (in fact, it has zero resistance!).

    As a more practical example, consider superfluids; they have no resistance, and are thus not ohmic, but we can still run steady currents (though extremely high) through them.

    The other statements made cannot be directly deduced from the propositions made in the problem.

    Good luck! :smile:
  14. Jan 2, 2009 #13
    Yes that makes sense, thanks.

    I never did chem at a-level.

  15. Jan 2, 2009 #14
    WOW great answer it seems perfectly logical. Thanks
  16. Jan 2, 2009 #15
    I found chem pretty boring and thats about the only knowledge ive used in physics so far so I wouldnt worry about not having done it.
  17. Jan 2, 2009 #16
    You're welcome :smile:
  18. Jan 2, 2009 #17
    Yes thank you very much.

    Do you agree with Besh's answer for part iv?

    I do however have some concerns over part v seeing as it seems you are over complicating the question. I'm thinking it may be answer e, due to the fact that I doubt they would introduce a non ohmic material at this level. Even stating it is a thin conducting wire.

    Last edited: Jan 2, 2009
  19. Jan 2, 2009 #18
    Yes, I arrived at the same answer as Besh. It is true that my answer is a bit complicated; this is why I am not completely sure. If this was an exam taken from a high school Physics C course or introductory physics course in a university, then my answer would be a bit too much, but then if it were the test were from say, a modern physics or higher-level electromagnetism course, then my answer would probably be correct. :smile:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Charged Particles
  1. Particle charge (Replies: 3)

  2. Charged Particles (Replies: 2)