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Homework Help: Electric Dipole Moment, Potential, and Field of contiuous charge

  1. Jun 17, 2013 #1
    1. The problem statement, all variables and given/known data

    There are three charged line segments, each with linear charge density λ, extending from the origin to (a,0,0), from the origin to (0,b,0), and from the origin to (0,0,c).

    a) Find the dipole moment of this charge distribution.
    b) Find the first two terms in the multipole expansion of the potential V(0,0,z) on the z axis for z≫a,b,c.
    c) What are the monopole and dipole contributions to E(0,0,z) for z≫a,b,c.

    2. Relevant equations

    ##\mathbf p = \int \mathbf x' \rho (\mathbf x')d^3x'##

    ##V(\mathbf x ) = \frac{1}{4\pi\epsilon_0} \Big[\frac{Q}{r} + \frac{\mathbf p \cdot \hat{\mathbf r}}{r^2}\Big]##

    ##\mathbf E(\mathbf x) = -\nabla V(\mathbf x)##

    3. The attempt at a solution

    This problem, as far as I can figure, should be fairly straightforward, but I just don't think I'm getting the correct answer (although I don't know what the actual answer is).

    For part a, I would think that it is as simple as integrating over the three line segments. Given that ##\mathbf x = x\hat{\mathbf i} + y\hat{\mathbf j} + z\hat{\mathbf k}## the dipole moment integral should turn into:

    ##\mathbf p = \int_0^a \lambda x \hat{\mathbf i} \, dx + \int_0^b \lambda y \hat{\mathbf j} \, dy + \int_0^c \lambda z \hat{\mathbf k} \, dz##

    ##\mathbf p = \frac{\lambda}{2}(a^2\hat{\mathbf i}+b^2\hat{\mathbf j}+c^2\hat{\mathbf k})##

    Unless I've made a grievous mistake, this is the dipole moment for this charge distribution as I see it.

    Part b then calculates the asymptotic form of the potential using the multipole expansion. The equation, using the first two terms is above. The total charge of the system is clearly given by ##Q=\lambda(a+b+c)##. The only thing that really needs any thought to calculate is the ##\mathbf p \cdot \hat{\mathbf r}##. Because the problem asks for finding the potential when ##\mathbf x = z\hat{\mathbf k}## that means that ##\hat{\mathbf r} = \hat{\mathbf k}## so that ##\mathbf p \cdot \hat{r} = \mathbf p \cdot \hat{\mathbf k} = \frac{\lambda c^2}{2}##. Using all this together the potential should be:

    ##V(0,0,z)=\frac{1}{4\pi\epsilon_0}\Big[\frac{\lambda(a+b+c)}{z} + \frac{\lambda c^2}{2z^2} \Big]##

    To calculate the electric field in part c, I can simply take the gradient of this function in Cartesian coordinates, which in this case is only the derivative with respect to z. I get:

    ##\mathbf E(0,0,z)=\frac{\lambda}{4\pi\epsilon_0}\Big[\frac{(a+b+c)}{z^2} + \frac{c^2}{z^3} \Big]\hat{\mathbf k}##

    I just want to make sure my thought process is correct for this problem. Did I do something wrong in my calculations or are these the correct answers. I ask because this problem is from Electromagnetism by Pollack and Stump and the question after this is very similar but with the line charges distributed differently. They do provide an answer for part b) in that case and if I follow the same line of logic as with this problem, I do not get the answer they do.
  2. jcsd
  3. Jun 17, 2013 #2


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    Hello zephyr5050 and welcome to PF!

    Your result for V(0,0,z) looks good to me.

    However, to find E(0, 0, z) you will need to find V(x, y, z) for nonzero (but small) values of x and y since you need the functional form of V in order to find the gradient of V.
  4. Jun 18, 2013 #3
    Thanks for the help TSny! I see what you're saying for the E(0,0,z). I had overlooked that point. I was fairly certain I was calculating at least the potential correctly, I just wanted a second opinion. Anyway, thanks.
  5. Oct 17, 2013 #4

    i have this same problem except the line charges are placed symmetrically around the three axis. that is they extend from -a,0,0 to a,0,0 from 0,-b,0 to 0,b,0 from 0,0-c to 0,0,c.
    need someone to work out the solution so i can see whats happening here.
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