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Homework Help: Some Electric Potential energy questions

  1. Feb 13, 2006 #1
    A charge of 3.19 µC is held fixed at the origin. A second charge of 3.19 µC is released from rest at the position (1.15 m, 0.530 m) .

    (a) If the mass of the second charge is 3.10 g, what is its speed when it moves infinitely far from the origin?


    ok so i've broken it up into x and y components and used pythagorean theorem to find the line that represents the force that the second charge exerts on the fixed charged [tex]\sqrt{1.15^2 + .530^2}[/tex] and my answer was 1.266 m. Then i used tan[tex]^-1[/tex] (.530/1.15) to find theta which came out to be 11.98[tex]^0[/tex]. Then i calculated 1.266 (cos 11.98) for the positive x-direction and 1.266(sin 11.98) for the positive y-direction and i got 1.2384 m for the x and .26278 m for the y. Then i plugged those values back into pythagorean theorem again to find the net force [tex]\sqrt{.26278^2 + 1.2384^2}[/tex] and got 1.2666N. Then i used the Force per unit charge to find the net electric field and got (1.2666/3.19E-6) = 3.97E5 and multiplied by 2 since there are 2 charges and got 7.94E5 as my net electric field. And then i plugged that value into [tex]\sqrt{2q\Delta V/m}[/tex] to find velocity. Obviously i'm doing something wrong somewhere because i'm not getting the right answer... any help would be greatly appreciated.
     
    Last edited: Feb 13, 2006
  2. jcsd
  3. Feb 13, 2006 #2
    What's the right answer ?

    Hint : what is the potential if the charge is that far away from the first charge.

    marlon
     
  4. Feb 13, 2006 #3
    well that's what i'm trying to figure out..
     
  5. Feb 13, 2006 #4
    Well,

    what's the forumula for the potential ?

    Again, what's the right answer ? You told in your first post that you keep getting the wrong answer ? How do you know this ?

    marlon
     
  6. Feb 13, 2006 #5
    formula for potential is [tex]\Delta U/q[/tex] but don't you have to figure out net electric field first since [tex]\Delta U = qEd[/tex] ??

    i'm using webassign so the answers i keep entering are wrong
     
    Last edited: Feb 13, 2006
  7. Feb 13, 2006 #6
    The formula for the potential of a charge q felt at a distance r is [tex]U = k \frac{1}{r}[/tex] where k is a constant that does not matter here. Now, what's the U if q is at spatial infinity ?

    It is important to understand the concept of potential (which is the aim of your exercise). A potential is actually potential energy per unit of electric charge. So two electric charges q1 and q2 have potential energy [tex]k \frac{q1*q2}{r}[/tex] where r is the distance between those two charges. If you devide this expression by, let's say, q1 you get the potential of q2. This potential, actually is a measure of how big the interaction (ie the magnitude of the potential energy) with an electrical charge will be if you put some charge q1 on a distance r from q2.



    marlon
     
    Last edited: Feb 13, 2006
  8. Feb 13, 2006 #7
    thank you for your help, i understand now..
     
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