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Homework Help: Finding Electric Fields

  1. Jul 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A point exists a distance D away from a rod of length L ( (point) |-----d-----| (ROD) ) So the point is in allignment with the rod, charged with a uniform charge density of -6.5uC/m

    2. Relevant equations

    E= Integral(k(dq/r^2)

    3. The attempt at a solution

    Im not asking for help on finding the answer, i just want to understand a bit more about charge density.

    It says that it is (-), in short, does that mean that the rod has a negative charge? meaning that the point (positive ) will be attracted to the rod? Thanks.

    One more thing, is the radius (r) the distance from the point to the center of the rod?
    Last edited: Jul 17, 2008
  2. jcsd
  3. Jul 18, 2008 #2


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    Hi NBAJam100,

    If the charge density is negative, then the charge will be negative. (When the charge density is uniform, the linear charge density is just how much charge there is per unit length.)

    Do you mean the r in your integral? That r is the distance from the dq to the point where you're calculating the field at.

    One way to understand the integral is to say we would like to break up the rod into an series of point charges, each with charge dq. The field from each of these charges dq is found from Coulomb's law (that's where the integrand comes from, and that's where the r comes in), and the integration adds up the fields from all of those charges.
  4. Jul 18, 2008 #3
    So if the r comes from the distance from the point to the dq, then does that mean that the r is not constant and has to be considered in the integral because there are many dq's on the rod (all of them make up Q). Or do you mean that I am measuring only one distance, from the point to that one sinlge dq that i am basing the problem off of? Thanks alph
  5. Jul 18, 2008 #4


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    The value r is not a constant and is part of the integration. You'll need to rewrite both dq and r in terms of an integration variable.

    So, for example, if you decide to integrate from the left end to the right end of the rod, and let the integration variable x represent the distance from the left end, what would dq and r be?
  6. Jul 19, 2008 #5

    Hmmm... ok i kind of think i got what you are saying. Either that or this is totally wrong:

    So i did E= k (integral) dq/x^2. Using the charge density (-6.5x10^-6 C/m) I found dq= (-6.5x10^-6 C/m)(dl). So i plugged that in to the integral and pulled the charge density out as a constant. After integration i got E= k(-6.5x10^-6 C/m) (-1/x)... now i think this is right, but im not sure what to plug in to get the electric field. The rod is 1.5m long and the point is .5m away from the rod. So would i plug in 2m (the total distance from the end of the rod to the point) to get the electric field at the point? If that is the case, i found the answer to be 29,250 N/C
    Last edited: Jul 19, 2008
  7. Jul 19, 2008 #6


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    This will be a definite integral, so you need to determine the limits.

    The integrand was Coulomb's law for each dq; so then what will be the limits of the integral? (Remember it will be different depending on how you defined your integation variable.)
  8. Jul 19, 2008 #7

    well, from the information i have i dont think i can determine the area of the rod... but for some reason i feel that the area should be a factor... regardless, if my integration variable was x, (or l for distance of length). So wouldnt the bounds be 0-2m. Or, 0m= the end of the rod, 2m = the end of the rod to the point. So by my logic, my answer would be correct. (assuming my logic is correct).
  9. Jul 19, 2008 #8


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    I don't know what the 2 meters corresponds to. I thought all we had was d=distance from point to rod, and L=length of rod?

    But about the limits: In the last post you had the integrand as:

    where [itex]\lambda[/itex] is the linear charge density of the rod. (There are other ways to set up the integrand, but this is fine.) Now you want to integrate this, and to do that you need to decide on limits.

    So your integration variable is x, and you want to integrate from the left end to the right end of the rod. At the left end, what is the value of x (in terms of d and L)? That will be the lower limit; in other words it is the value of the integration variable x that makes the integrand correct for the dq at the left end.

    Then the upper limit will be the value of x will correspond to the right end of the rod.

    About your post: I don't know what the 2 meters is. However, with the way you set up your integrand, your lower limit cannot be 0. Your denominator is x^2, so x is the distance from dq to the point. But the left end is not zero meters away, so the lower limit can't be 0. Do you see what your two limits have to be?
  10. Jul 20, 2008 #9

    Ok, so d= .5m (the point to the beginning of the rod), L of the rod = 1.5 meters. So, my lower bound would be .5 (or the distance from the point to the rod (d)), my upper bound would be 2m, the distance from the point to the end of the rod (d+L). So i would just plug those values in to get the correct value...
  11. Jul 20, 2008 #10


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    Those sound like the right integration limits to me. Did you get the right answer?
  12. Jul 20, 2008 #11
    For my final answer i got E= -87,750 N/C...

    Thanks for all of the help alph! i really appreciate it!
  13. Jul 20, 2008 #12


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    That seems like the right magnitude to me.

    But you might need to be careful: what does the minus sign mean in your answer?

    If you include a sign then it indicates the direction of this electric field; so for this case, which way is this electric field pointing at that point? And does that match with your definition of positive and negative directions?
  14. Jul 20, 2008 #13
    I determined that the point was positive, and the rod was negatively charged (found by the charge density being negative). So if the point is positive, and the rod is negative, then the point is attracted to the rod. So the negative sign is right, the direction of the field from the point would be pointing towards the rod (or towards the negative (-)). Correct?
  15. Jul 20, 2008 #14


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    No, I don't believe that's correct. The positive or negative sign of the electric field will tell the direction--and that depends on the coordinate system. So if you set up your x-axis in the normal way with positive meaning to the right and negative meaning leftwards, then a negative value of the electric field would mean that the electric field is pointing to the left. But you said it's pointing toward the rod, which is to the right.

    Think of it this way: you found the electric field at a point to the left of the rod, and you know the electric field there is pointing to the right (toward the rod).

    Now suppose you next found the electric field at the opposite side of the rod (on the right), the same distance away. It would have the same strength of the electric field, and it's still pointing toward the rod, but now "toward the rod" means to the left. Since it points in an opposite direction, it would need to have the opposite sign of your original point. So the sign of the electric field is not the sign of the charge, it's simply indicates the direction of the field.

    (Remember that what you are really doing with the sign is reporting the x-component of the electric field, because you already know it has to be in the x-direction.)
  16. Jul 20, 2008 #15

    Ok, so i THINK i get what this means... are you saying that i was finding the direction of the field from the left of the point... It was pretty much a given that from the right side of the point the field is facing towards the rod... but what that sign indicated was that the sign is in the negative direction from the left side of the point? If so, i understand that fully... So pretty much it would look like this:

    ---> = field lines

    <---<---<---( + )--->--->---> ( ( - ) Charged rod) <---<---<---<

    And then there would be other field lines curving from the outside of the positive going to the negative.
    Once again, i cant thank you enough for being so patient with me alph.
    Last edited: Jul 20, 2008
  17. Jul 20, 2008 #16


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    I'm a bit confused about your diagram. There's not a charge at the point you are calculating is there? Usually they say they want you to find the electric field at a certain point; but there's nothing actually at that point. It's just empty space.

    But even if there is a charge there (which it really does not sound like there is), if you're asking about the electric field of the rod, the answer will be the same whether there's a positive charge, negative charge, or nothing at that point. You are just asked what is the electric field from the rod at that point, right? So the diagram is:

    --->--->---> ( ( - ) Charged rod) <---<---<---<-

    By the way, I noticed you never explicitly stated the question. Usually these questions have two forms: 1. they ask for the magnitude of the field, in which case just the absolute value is needed; or 2. they want the magnitude and direction of the field, in which case the answer here would be "87750 N/C to the right" or "87750 N/C in the positive x-direction" or something similar.

    Usually, if they wanted the sign of the number to indicate the direction, they would say that explicitly (something like "in your answer, let positive be to the right"). So if it's still not making sense, would you post the exact statement of the question?
  18. Jul 20, 2008 #17
    Ok, im getting two things mixed up i think haha. You're right, there is no charge there, the question is this: "Find the magnitude and direction of the net electric field at point P, a distance of .5m away from the end of a rod of length 1.5m charged with a uniform charge density of -6.5 uC/m." I understand how to get the answer, i guess i was just confused about the negative in my answer with regards to the direction of the field.
  19. Jul 20, 2008 #18


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    Oh, okay. So the magnitude would be just the 87750 N/C, and the direction would be to the right, like we had a few posts back. Looks great!
  20. Jul 21, 2008 #19
    Thank a million alph! i must say i learned a hugeee amount from these posts!
  21. Jul 21, 2008 #20


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    Sure, glad to help!
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