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Homework Help: Electric Field of a Thin Rod

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Thin Rod
    - Length = 5.0cm
    - Total Charge = 8.4 nc

    Calculate the electric field at point p, 3cm away.
    2. Relevant equations

    E = (1/4∏ε) * (q/r^2)

    λ = Q/L

    3. The attempt at a solution

    I think you treat the rod as a series of point charges, meaning the electric field would be the sum of the individual electric fields.
    If the rod was vertical, I would know what to do as you have to calculate the integral due to the varying distances and the charges are to the point using trigonometry and Pythagoras, we went through an example of this.
    In this case the rod is horizontal, but obviously the distance for the 'point charges' still varies so I've been trying to construct an integral for this.

    E = (1/4∏ε) * Ʃ(Δq/d^2_i)

    I've also calculate the charge density.

    Attached Files:

  2. jcsd
  3. Feb 20, 2012 #2
    You could set the point p as the origin and integrate the electric field due to the charges, it should look something like this:
  4. Feb 20, 2012 #3


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    Gold Member

    Right. This is the (classical) principle of superposition. The E field for many charges is the (vector) sum of the E fields for each. Then in the limit this sum becomes an integral.

    You replace summing over charges [itex]q_k[/itex] with integrating over the differential charge, [itex] dq=\rho dV[/itex]. (in this case with a long thin distribution, [itex]\rho dV =\rho A dx = \lambda dx[/itex] where [itex] A[/itex] is cross section area, or equivalently [itex]\lambda = \rho A[/itex] is density of charge per unit length.

    All that remains then is to use Coulomb's law to express the differential field [itex]dE[/itex] for a differential charge [itex]dq[/itex] at its given position and integrate.
  5. Feb 21, 2012 #4
    I still don't think I get if I'm honest.

    Would I get...

    E = (1/4ε) * ∫ (dq/d^2_i)
    with the limits of the minimum and maximum distance from point p

    Then I would need to substitute in something for 'dq' or 'd'?
  6. Feb 21, 2012 #5
    You're missing a π in the constant out front.
    More importantly, you need to figure out dq in terms of dx. In this case, it's simply λdx.
  7. Feb 21, 2012 #6
    Then the λ comes out because it is a constant right?
    The missing ∏ was just a typo.

    E = (1/4∏ε) * ∫ ( λdx/x)
    E = (1/4∏ε) * λ ∫ ( dx/x)
    E = (1/4∏ε) * λ [ln(x)]

    Then sub limits in?

    Cheers btw mate.
  8. Feb 21, 2012 #7
    Close, the integrand is dx/x2.
  9. Feb 21, 2012 #8
    Oh yeah forgot it was squared on the bottom.

    I have another question as well.

    If I have three charges in a line and there is one above them and I want to find the net force exerted on them by the charge above.
    Do I find the individual force on each , and then combine these forces?

    The charges are arranged like this

    Okay trying to draw a diagram didn't really work.
    But the shape forms a triangle.
    A negative charge at the top.
    A positive charge in each of the corners.
    Then a negative charge inbetween the two positive charges.
  10. Feb 21, 2012 #9
    Does my interpretation of this triangle of charge set up make sense?

    The four charges are positioned as follows: at the origin you have a negative charge; at +d and -d (on the x-axis) you have positive charges; and at +r (on the y axis) you have a negative charge.

    Is that the correct set-up?
  11. Feb 21, 2012 #10
    Yeah that is it.
    Don't know why I didn't just give it in terms of coordinates, that is how I have it wrote out in front of me.
  12. Feb 21, 2012 #11
    You need to write an expression for the force on the negative charge located at (0, r). To do this you have to add up the forces that each of the other charges exerts on this charge.

    Recall Coulomb's Law, which states:

    F[itex]_{coul} = \frac{1}{4\pi\epsilon_{0}} \frac{q_{1}q_{2}}{r_{12}^{2}}[/itex]

    Where [itex]q_{1}[/itex] and [itex]q_{2}[/itex] are the charges [for this question one of these charges will always be the charge located at (0,r)] and [itex]r_{12}[/itex] is the distance between the first charge and the second charge; this is not always equivalent to the r I have used in describing the position of the charge at (0, r).

    Then recall that you can take the vector sum of the three forces you find and you will have an expression for the net force on the charge at (0,r).
  13. Feb 21, 2012 #12
    Actually one more question, are the limits 0.5 and 5.5?

    As in the max distance the charge is and the shortest distance the charge is from the point?
  14. Feb 21, 2012 #13
    The lower bound of integration should be the shortest distance from your line of charge to the point you are interested in; the upper bound should be the largest distance from your line of charge to the point you are interested in.

    This means your lower bound should be the distance between the end closest to the point and the upper bound should be the distance from the far end to the point.

    If the point you are interested in determining the electric field at is at 0.5, the close end is at the origin and the far end is at -5 then those are the correct limits of integration.

    EDIT: I looked at the attached picture and your limits of integration are correct!
  15. Feb 21, 2012 #14
    What I did was use
    F[itex]_{coul} = \frac{1}{4\pi\epsilon_{0}} \frac{q_{1}q_{2}}{r_{12}^{2}}[/itex]

    Then I calculated the force that the charge at (0,2) would apply to charges at (-1,0),(0,0) and (1,0) individually
    Using the appropriate r in each case, which was √5 (Found using trig), 2, and √5 again.

    So I now have the force that each of the charges would experience due to the charge at (0,2), but it asks for the net force on them, so I though I would have to combine them somehow, or is that all wrong?
  16. Feb 21, 2012 #15
    I may have misinterpreted the question incorrectly.

    Does the question want you to find the net force on each of the three charges or does the question want you to determine the net force on the charge at (0,r) due to the three charges?

    Either way, if you are looking for the net force you need to consider the force from each of the other three charges acting on the charge you are interested in.
  17. Feb 21, 2012 #16
    Sorry I haven't made it very clear,
    It asks you to find the net charge on the three charges on the line y=0 due to the charge at (0,2).
    I have the force on each charge, but I'm not sure how to make them into the net force?
  18. Feb 21, 2012 #17
    It seems poorly worded. To find a net force you must consider all the forces.

    For example, consider the charge located at the origin. To determine the net force on this charge you would consider the force acting on this charge from the charge located at (-d,0), the force acting on this charge from the charge located at (d,0) and the force acting on this charge from the charge at (0,r). You would find three forces and use vector addition to find the summation which gives you the net force.

    If the question asks for the "net (I think you mean) force due to the charge at (0,2)" then the net force due to this one charge at (0,2) is simply the one expression for Coulomb's Law.

    If it asks for the net force I am inclined to consider the force from all three of the other charges.

    If this is for a problem set or homework assignment I suggest you e-mail your Professor or TA for clarification as to which one it is. If it is a practice problem to which the solution is provided, try it both ways to see which one is correct (which one its asking for). If no solution is provided, your guess is as good as mine (the question seems ambiguous and poorly worded).
  19. Feb 21, 2012 #18
    I've attached the actual question, in case it is the way I am explaining it.

    Attached Files:

  20. Feb 21, 2012 #19
    Okay, so the question is asking for the net force exerted by these charges on the charge located on the y-axis.

    If we call the three charges on the axis 1, 2, and 3 where 1 is located at (1,0), 2 at (0,0) and 3 at (-1,0) and the last charge, 4 at (0,2) we need to determine the Coulomb force between 1 and 4, 2 and 4, and 3 and 4.

    Using Coulomb's Law we will find three expressions. You will then have to rewrite these expressions in terms of x- and y- components and add all the x-components together and then all the y-components together. By symmetry show that the net force in the x direction is 0.

    Once you do this you should find the net force has only a y-component and this will be your net force that the question is asking for.
  21. Feb 21, 2012 #20
    Ah I see, I went wrong with it pretty quickly then as I didn't divide it into the components.

    So, let me just check this with you
    X - Components

    (F2)x = (1/4∏ε) *[ ((Q2)x*(Q4)x)) / 0^2 ] = 0 as over 0
    (F1)x = (1/4∏ε) *[ ((Q1)x*(Q4)x)) / 1^2 ]
    (F3)x = (1/4∏ε) *[ ((Q3)x*(Q4)x)) / 1^2 ]

    - The last two cancel because of symmetry

    Y- Components

    (F2)y = (1/4∏ε) *[ ((Q2)y*(Q4)y)) / 2^2 ]
    (F1)y = (1/4∏ε) *[ ((Q1)y*(Q4)y)) / 2^2 ]
    (F3)y = (1/4∏ε) *[ ((Q3)y*(Q4)y)) / 2^2 ]

    Then add the answers to these three equations together?
  22. Feb 21, 2012 #21
    I have never done components that way. The best way is to find the force first and the split the force into its components. I'll do an example for the force between the charge at (-d, 0) and the charge at (0, r) (I will call them [itex]q_{1}[/itex] and [itex]q_{2}[/itex] respectively).

    For your problem d = 1, r = 2, [itex]q_{1}[/itex] = 3e(-4) C, [itex]q_{2}[/itex] = - 3e(-5) C.

    [itex]\textbf{F}_{Coul} = \frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r_{12}^{2}}[/itex]

    [itex]r_{12}[/itex] is the distance between [itex]q_{1}[/itex] and [itex]q_{2}[/itex]. We can express this distance by drawing a triangle; you should find the following:

    [itex]r_{12}^{2} = 1^{2} + 2^{2} = 1 + 4 = 5[/itex]

    Knowing this you should be able to express the x- and y-components of the Coulomb force for this pair of charges. Then you simply have to do the same thing for the other two pairs of charges. Once you have done this you must add all the x-components and y-components together.

    So we can write:

    [itex]\textbf{F}_{Coul} = \frac{1}{4\pi\epsilon_{0}}\frac{(3x10^{-4})(-3x10^{-5})}{5}[/itex] (the x here is not a variable, it is a multiplication symbol).

    [itex]\textbf{F}_{Coul} = k_{e} \frac{(3x10^{-4})(-3x10^{-5})}{5}[/itex], [itex]k_{e} = 8.99 x 10^{9}[/itex] is Coulomb's constant .

    So you should find [itex]\textbf{F}_{Coul} = -16.182 N[/itex] which means that the force [itex]q_{1}[/itex] exerts on [itex]q_{2}[/itex] is has magnitude 16.182 N; its being negative suggests that it is an attractive force (opposite charges).

    Now that we know the force we can determine the components of this force. If you were to draw a triangle with this value of [itex]\textbf{F}_{Coul}[/itex] as the hypoteneuse the other two sides could be called [itex]F_{x}[/itex] and [itex]F_{y}[/itex] for the x- and y-components of [itex]\textbf{F}_{Coul}[/itex], respectively.

    To determine these values you must take the sine or cosine of an angle; we will call this angle [itex]\theta[/itex]. This triangle will be similar to the triangle I have described above in determining [itex]r_{12}[/itex].

    IMPORTANT: In your method you have done something fundamentally wrong; you have looked at the distance between points r as the distance in only the x- or y-direction. This is incorrect. For example, if you look at the term you have for the x-component of F2 (the force acting on the charge located at (0,1) from the charge located at (0,0) you have used r = 0. This would make F be infinite! This is not the case.
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