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Some of my Winter Break physics problems

  1. Dec 29, 2007 #1
    [SOLVED] Some of my Winter Break physics problems

    Hey guys, I'm new to this site, and as the way things are going with my physics class, this won't be my last request for help lol.

    Well anyway, our teacher assigned us a few problems to do over break, and since I have a week left I decided to start doing them now..

    The first problem looks like this:

    -------------------------------------------------------
    Two point charges q1 = +2.20 nC and q2 = -6.50 nC are 0.100 m apart. Point A is midway between them and point B is 0.080 m from q1 and 0.060 m from q2 (Fig. 23.29). Take the electric potential to be zero at infinity.

    (a) Find the potential at point A. **right**
    (b) Find the potential at point B. **right**
    (c) Find the work done by the electric field on a charge of 2.50 nC that travels from point B to point A.

    -----------------------------------------

    I got A and B right, but C I've got no clue how to do.. I was going over my notes and saw some sort of integration stuff, but I can't figure out how to do any of that!

    ================================

    A total electric charge of 2.10 nC is distributed uniformly over the surface of a metal sphere with a radius of 36.0 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere.

    (a) 48.0 cm
    (b) 36.0 cm
    (c) 12.0 cm

    -----------------------------------------
    This one, I've got NO CLUE how to do at all...
    I know the basic equations v=U/q but what do you plug in where!?

    -----------------------------------------

    I guess I'll just leave it at these two for now until I get these right...

    Thanks alot!
    Peter
     
  2. jcsd
  3. Dec 29, 2007 #2

    rock.freak667

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    What is the potential difference between B and A? What does the product of the potential difference and the charge give you?
     
  4. Dec 29, 2007 #3
    For first part use equation for points (A and B) V = Kq/r for Work = qV = U, so it will be q(Vb-Va)

    For part two use Intergral of E dot dr should be equation in book
     
    Last edited: Dec 29, 2007
  5. Dec 29, 2007 #4
    I don't know, do I do it the same way as i did for A and B?

    How do you figure out the distance though?

    Here's basically what I have on paper..
    [​IMG]
     
  6. Dec 29, 2007 #5

    rock.freak667

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    Well if you have the potential at A,[itex]V_A[/itex], and the potential at B,[itex]V_B[/itex]. What is the potential difference? when you get that all you need to do is multiply by the charge that you moving from B to A
     
  7. Dec 31, 2007 #6
    Okay thanks alot!
    First problem is completely finished

    But the second one I don't get..
    I looked up the equation you were talking about
    W(ab) = Integral of E(dot) dr

    There's an example with something to do with infinity and r/2, for the first one it's zero, and the second one is r/2 so do i divide the radius by 2 and then multiply it by the what?

    I'm sorry..
     
  8. Dec 31, 2007 #7
    Use gauss law to find E, the "r/2 and infinity" are just different limits of integration.
     
  9. Jan 1, 2008 #8
    Yeah I'm sorry but I don't get it..
    E would be Q/4*pi*E(naught)
    right?
    But what do you multiply that by??
     
  10. Jan 3, 2008 #9
    Your E is wrong missing R squared.
    "But what do you multiply that by??" how do you simplify a dot product? (hint: you can choose any path for dr, and you know how the vector field E looks like)
     
  11. Jan 4, 2008 #10
    oh woops yeah I did forget R squared
    so for vector path dr I'd put .48m because that's what they're looking for?
    So .48Q/4piEnaughtRsquared?

    Sorry, I'm just terrible at gaussian surfaces.. I never really understood how to do them
     
  12. Jan 4, 2008 #11
    For this case dr it is just the integration operator. And since you want from V(0m) - V(.48m) = integral of E dot dr, your limits would be 0 -> .48m (you might have to split it into two integrals, since you cannot have 1/0)
     
    Last edited: Jan 4, 2008
  13. Jan 6, 2008 #12
    i got it!
    thanks alot!!
     
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