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Homework Help: Finding Electric Field on the Bisector of a Non-Uniformly Charged Rod

  1. Dec 31, 2009 #1
    This question has been giving my brain some trouble. I think i have the Electric Field integral calculated correctly (but I'll post that up here too just in case I'm wrong which is likely) but when I get down to finding the charge on the rod I literally am making a wild guess at the solution.

    1. The problem statement, all variables and given/known data
    A rod with length L has a charge density of Ax where A is a constant and x is measured from the center of the rod, with positive to the right. Set up the integral to find the electric field at point P (which is at a distance h above the center of the rod) but do not evaluate. Be sure to tell the direction of the net electric field. Also find the total charge on the rod.

    2. Relevant equations
    dEp = kdq/r2 where k = 8.99E9

    3. The attempt at a solution
    So for the integral, I set dq = Ax*dx and r = [tex]\sqrt{L^2/4 + h^2}[/tex] I stated that, due to symmetry, the net electric field on the x-axis should be equal to 0, though I don't know if I can make that assumption since I don't know the charge on the rod? Regardless, I ended up getting this integral:
    Ep = kA/(L[tex]^2[/tex]/4 + h[tex]^2[/tex]) [tex]\int { xdx}[/tex] in the +y direction. The integral is from 0 to h, as well. Not sure if those are the correct bounds to be using, but it made sense to me.

    From there I figured the only way I could calculate the charge is by solving the integral and then solving for Q:

    Ep = kAx[tex]^2[/tex]/(L[tex]^2[/tex]/2 + h[tex]^2[/tex]/2) from 0 to h

    Ep = 2kAh[tex]^2[/tex]/(L[tex]^2[/tex] + h[tex]^2[/tex]) and since [tex]\lambda[/tex] = Ax and the units for [tex]\lambda[/tex] = N/C you can figure out that the units for A must be C/m[tex]^2[/tex] so A = Q/L[tex]^2[/tex]

    Ep = 2kQh[tex]^2[/tex]/(L[tex]^4[/tex] + h[tex]^2[/tex]*L[tex]^2[/tex])

    L[tex]^4[/tex] + h[tex]^2[/tex]*L[tex]^2[/tex] = 2kQh[tex]^2[/tex]

    Q = L[tex]^4[/tex] + h^2[tex]^2[/tex]*L[tex]^2[/tex]/2kh[tex]^2[/tex]


    So I am almost 100% sure that the way I solved for charge will make any physicist cry because, honestly, it just looks wrong. Besides, I believe that the way I solved this here also assumes that the Electric Field at P is equal to 1. If anyone can help me through this problem and help me understand my mistakes I would greatly appreciate it. It's weird that this problem is giving me trouble when other things like charged arcs are easy for me to grasp.

    Thanks in advance!

  2. jcsd
  3. Dec 31, 2009 #2


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    Well there's your (first) problem :wink: dq is a charge. A is a charge density. To get a charge from a charge density, you need to multiply by... what kind of quantity?

    Here's how you could (should) have caught that mistake: what are the units of Ax*dx? Do they agree with the units of dq? They must, if you're going to set them equal to each other.
    Remember that r is always the distance between the charge and the point (P) where you're computing the field. You understand that a point at the end of the rod is further from P than the point at the center of the rod, right? So the point at the end of the rod has a larger r than the point at the center. That tells you that r has to be variable - it should depend on x. Based on that, can you come up with an expression for r, the distance between P and a point on the rod, given x for that point on the rod?
  4. Dec 31, 2009 #3
    Ok, here's my logic behind the dq thing first- For uniformly charged rods dq is set equal to [tex]\lambda[/tex]ds where ds = rdx. So for this problem since [tex]\lambda[/tex] = Ax, dq should therefore be equal to Axrdx, correct?

    Now for my "r" statement, am I on the right track at least? Should x be there in place of the square root expression? Since that is just a distance from point P to the end of either side of the rod and that distance is what should be variable, am I right, then, in assuming r = x?
    Last edited: Dec 31, 2009
  5. Dec 31, 2009 #4
    I would have assumed Ax is the linear charge density (C/m), otherwise you would need the dimensions of the rod
  6. Dec 31, 2009 #5
  7. Dec 31, 2009 #6
    Oh, I seem to have forgotten to add the word "Linear" in front of "Charge Density" in the copy. It's Linear Charge Density, or [tex]\lambda[/tex]

    "h" is representative of a certain height that point P is located above the center of a horizontal rod. Crude image:

    Code (Text):

               |<- h
     ================== <- Rod
    |---------- L ----------|
  8. Dec 31, 2009 #7


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    No! Again, think about units: What should the units of ds be? (length) What are the units of rdx?

    That will at least tell you that ds = rdx is wrong, but I'm thinking you may not know how to figure out what the right answer is. ds is supposed to be a differential element of length along the distribution of the charge. What variable measures length along the distribution of the charge - that is, along the rod? That variable is what takes the place of s in this problem.
    ooops, I misread the problem. I thought A was the charge density - silly me.
    Sorry, actually you were right the first time: dq = Axdx. I should have read the problem more carefully.

    x should be in the square root expression, but not in place of it. r is supposed to be the distance from point P to a point on the rod, but it is not equal to x. Think about this: at the center of the rod, x = 0. What is r at that point?

    Once you figure that out, same question for a point off the center of the rod, where x is not zero: what is r, the distance from point P to that point?
  9. Dec 31, 2009 #8
    WHOOPS! I was looking at the wrong problem in my notes. ds should be equal to dx. I was looking at my charged arc problem where ds = rd[tex]\theta[/tex] and just didn't even realize it. For a second there I thought I was going crazy! Thanks for pointing that out!

    So r at x=0 is equal to h. So that means I need to have a variable distance across the rod which lies on the x-axis in order to figure out the length of "r" at each point on the rod. Thus, r should equal [tex]\sqrt{x^2 + h^2}[/tex] correct? My distance from point P on the y-axis doesn't change where as my distance on the x-axis does so this expression covers both those.

    Still, even if I am right, calculating that charge still seems odd. Assuming I have the correct integral would I solve it and then algebraically derive the value of Q in terms of h and L? Or am I WAY over-thinking the problem (which, nine time out of ten, is the case)?
  10. Dec 31, 2009 #9
    Why would h affect the total charge on the rod?

    How would you go about finding the charge on the rod if the linear charge density was constant? Thinking about that might help you set up an integral to find the charge when it isn't constant
  11. Dec 31, 2009 #10


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    Cool, glad we got that sorted out :wink:
    Yeah, you're overthinking it. Finding the charge is a separate calculation from finding the electric field. Use JaWiB's hint.
  12. Dec 31, 2009 #11
    That clear the dimension and the location of P up. But I still have problem if it is linear( 1 dimension) charge density. If you look at it in the way the charge is proportion to the distance from the center of the rod, you have to look at it as layers of thin cylinders that make up the rod and you have to take each point around the ring of radius of each cylinder to P to calculate the E. Seems you have to have a volume charge density and you integrate the E of each dV around the radius and along the L of the rod. [tex]dV=rd\phi dLdr[/tex]. And you have to calculate the E of each thin cylinder that make up the rod.

    I have not work this out and I have a few beers!!!! But the information given still border me!!!

    Happy New Year.
    Last edited: Dec 31, 2009
  13. Dec 31, 2009 #12


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    And this, folks, is why you shouldn't drink and derive :rolleyes: all together now... *groan*
  14. Dec 31, 2009 #13
    No you don't need a volume charge density, and you don't need to integrate around a ring. To find the electric field at P, find the contribution from an infinitesimal segment of the (1-dimensional) rod and integrate over the length of the rod.

    Perhaps you should wait until you're sober to figure this out :)
  15. Dec 31, 2009 #14
    Like how?!! The question is charge in a rod which is a 3 D object that have length L and a radius. Read the question, it said the charge at distance x from the center of the rod is Ax. This mean the rod is made up of rings of constant charge density rings. "ROD" tell you it is 3D object. This is not a line charge. Unless specified, center of the rod is axis line that go through the length of the rod where the radius equal to zero.

    On top of that, you have to have the dimension of each charged line inside the rod or else how do you calculate the total number of lines inside a 3D rod that have a volume. This is a 3D object, not a line charge. You just cannot have a 1 dimension line( no radius, no volume) to some how make up the volume of the rod of length L and radius r.

    Look at it another way, if each line charge have no cross section area, you can fit infinite number of charge lines into the rod, how do you define the charge. You can claim you can fit 1000 line or you can claim able to fit 10000000000000 line in the rod. You have to define in 3D which mean volume charge.
    Last edited: Jan 1, 2010
  16. Dec 31, 2009 #15
    Also the question never specify the distance "h" is much larger than the radius of the rod, so even at any point along L of the rod, the cross section of the rod at that point is an area of the circle. The distance from different points within the area of the circle to "P" is different, you cannot assume the rod is a line charge.

    This kind of question is common in electro static chapter of the EM books!!!! Until the original poster clarify everything, don't assume anything.
    Last edited: Jan 1, 2010
  17. Jan 1, 2010 #16
    can you give a more detail drawing, specify where is the x axis, y axis, r and h. From your original post #1 you said x is from the center of the rod, convensional thinking is you meant x is along the radius of the rod perpendicular to the length L of the rod. center of the rod always meant the axis of the rod( the line going through the the length of the rod and the center where you measure the radius of the rod. I hope you don't call center of the rod to mean the mid point of the length of the rod!!!!

    But in either case, you still have to deal with the cross sectional area where distance from different point of the cross sectional area to "P" is different, you cannot use line charge calculation. UNLESS you specify "h" is >> radius of rod, then you can do that.
    Last edited: Jan 1, 2010
  18. Jan 1, 2010 #17
    It is linear. It is given in the problem as linear. The X-axis is parallel to L and point P is located a distance "H" above the bisector of the rod, perpendicular to L, on the y-axis.

    @diazona/JaWib- I really have no idea how to calculate the charge for this object. I keep coming back to Gauss' Law but we've only just started using it and haven't moved past spheres yet and this seems much more complicated than anything I could handle so I know there must be another way but I'm not sure. My only other thought was to treat the rod like a point charge and use the equation E = kq/r^2 but I'm not sure if I can do that or why it would even work if I could.

    Wait, could this work? Could I say that since Ax is the linear charge density, it should be equal to Q/L and then say that Q is therefore equal to AxL? Or is that just circular logic and not really helping me find the charge at all? (I'm not even sure if I should be getting variables or actual numbers here)
  19. Jan 1, 2010 #18
    Thanks for the clarification. So this is how I read it:
    1) The rod is assumed to be a charge line with no radius so the charge density is represented as line charge which is linear along x axis.
    2) The charged line is from x=L/2 to x=-L/2 with the mid point at x=0.
    3) Point P on y axis where x=0 and P at y=h.
    4) at x=L/2 the line charge density is Ax which is positive.
    5) at x=-L/2, charge is negative.

    With that.
    1) you cannot use gauss law. Gauss only good for if L>>>h, AND the charge is constant through out the entire line. In another word, Gauss is only good if the resultant E field has only y component, the x component has to cancel out, no exception.
    2) You calculate the electric field at P by integrating the field contributed by each segment dx in the rod.
    3) the total charge of the rod should be zero!!!! Since the positive charge on the x>0 is equal and opposite to the negative charge on x<0!!!

    tell me if I interpret this correctly.
    Last edited: Jan 1, 2010
  20. Jan 1, 2010 #19
    Yup, you interpreted it correctly! And now I understand it. I was completely ignoring the statement "X is measured from the center of the rod with positive to the right". Man, reading the question over again would've saved me a lot of head scratching- thanks!
  21. Jan 1, 2010 #20
    Yep, That's why I wrote so much to clarify everything. I don't know exactly what class is this so I assume this is the first EM class and they are very specific on the terms, whether it is a line charge, a rod with diameter and the coordinates. My books never use linear charge density that really throw me off.

    Don't feel bad, I read the question at least 10 times last night and still not sure what it meant.
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