Electric Dipole Moment, Potential, and Field of contiuous charge

  • Thread starter zephyr5050
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  • #1
zephyr5050
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Homework Statement



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.

Homework 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)##

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.
 

Answers and Replies

  • #2
TSny
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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.
 
  • #3
zephyr5050
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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.
 
  • #4
hochstrasser
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help

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.
 
  • #5
theshape89
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I'm resurrecting a zombie thread here, but I think the answer given for the dipole potential is incorrect. TSny provides the correct method for obtaining it, despite saying that OP's answer for the dipole potential is correct. To expand a little, if we find ##V_{dipole}(x, y, z)## we get
\begin{align}
V_{dipole}(x, y, z) &= \frac{\hat{\textbf{r}} \cdot \textbf{p}}{r^{2}} \\
&= (\hat{\textbf{x}} + \hat{\textbf{y}} + \hat{\textbf{z}}) \cdot \frac{(\frac{1}{2} \lambda a^{2} \hat{\textbf{x}} + \frac{1}{2} \lambda b^{2} \hat{\textbf{y}} + \frac{1}{2} \lambda c^{2} \hat{\textbf{z}})}{x^{2}+y^{2}+z^{2}}
\end{align}
Setting ##x## and ##y## to ##0## to get ##V(0, 0, z)## then gives a different result to OP's solution.
 
Last edited:
  • #6
haruspex
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I'm resurrecting a zombie thread here, but I think the answer given for the dipole potential is incorrect. TSny provides the correct method for obtaining it, despite saying that OP's answer for the dipole potential is correct. To expand a little, if we find ##V_{dipole}(x, y, z)## we get
\begin{align}
V_{dipole}(x, y, z) &= \frac{\hat{\textbf{r}} \cdot \textbf{p}}{r^{2}} \\
&= (\hat{\textbf{x}} + \hat{\textbf{y}} + \hat{\textbf{z}}) \cdot \frac{(\frac{1}{2} \lambda a^{2} \hat{\textbf{x}} + \frac{1}{2} \lambda b^{2} \hat{\textbf{y}} + \frac{1}{2} \lambda c^{2} \hat{\textbf{z}})}{x^{2}+y^{2}+z^{2}}
\end{align}
Setting ##x## and ##y## to ##0## to get ##V(0, 0, z)## then gives a different result to OP's solution.
You seem to be saying ##\hat {\textbf{r}}=\hat {\textbf{x}}+\hat {\textbf{y}}+\hat {\textbf{z}}##.
 
  • #7
kuruman
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You seem to be saying ##\hat {\textbf{r}}=\hat {\textbf{x}}+\hat {\textbf{y}}+\hat {\textbf{z}}##.
Yes, that is a problem. The spatial dependence of the dipole potential is $$V_{\text{dip}}=\frac{\mathbf{r}\cdot\mathbf{p}}{r^3}=\frac{xp_x+yp_y+zp_z}{(x^2+y^2+z^2)^{3/2}}.$$It correctly predicts that the potential has odd parity, ##V_{\text{dip}}(-x,-y,-z)=-V_{\text{dip}}(x,y,z).## The substitution of ##\mathbf{\hat r}\cdot \mathbf{p}## for ##\mathbf{r}\cdot\mathbf{p}## forces the dipole potential to have even parity.
 
  • #8
TSny
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The substitution of ##\mathbf{\hat r}\cdot \mathbf{p}## for ##\mathbf{r}\cdot\mathbf{p}## forces the dipole potential to have even parity.
I'm not following this. Both ##\mathbf{\hat r} \cdot \mathbf{p}## and ##\mathbf r \cdot \mathbf{p}## have odd parity.

Using ##\mathbf{r} = r \mathbf{\hat r}##, we have$$V_{\rm dip}(\mathbf{r}) = \frac{\mathbf{r} \cdot \mathbf{p}}{r^3} = \frac{\mathbf{\hat r} \cdot \mathbf{p}}{r^2}.$$
 
  • #9
kuruman
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Using ##\mathbf{r} = r \mathbf{\hat r}##, we have$$V_{\rm dip}(\mathbf{r}) = \frac{\mathbf{r} \cdot \mathbf{p}}{r^3} = \frac{\mathbf{\hat r} \cdot \mathbf{p}}{r^2}.$$
Yes, but if you use ##\mathbf{\hat r} \cdot \mathbf{p}## to get ##V(x,y,z)## as required by the problem, $$\mathbf{\hat r} \cdot \mathbf{p}=(\mathbf{\hat x}+\mathbf{\hat y}+\mathbf{\hat z})\cdot(p_x\mathbf{\hat x}+p_y\mathbf{\hat y}+p_z\mathbf{\hat z})=p_x+p_y+p_z$$ so that $$V_{\text{dip}}(x,y,z) = \frac{p_x+p_y+p_z}{(x^2+y^2+z^2)}.$$This has even parity. If you use ##\mathbf{ r} \cdot \mathbf{p}##, you get the expression in post #7 which has odd parity.
 
  • #10
TSny
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Yes, but if you use ##\mathbf{\hat r} \cdot \mathbf{p}## to get ##V(x,y,z)## as required by the problem, $$\mathbf{\hat r} \cdot \mathbf{p}=(\mathbf{\hat x}+\mathbf{\hat y}+\mathbf{\hat z})\cdot(p_x\mathbf{\hat x}+p_y\mathbf{\hat y}+p_z\mathbf{\hat z})=p_x+p_y+p_z$$

##\mathbf{\hat r} \neq \mathbf{\hat x}+\mathbf{\hat y}+\mathbf{\hat z}##

The vector ##\mathbf{\hat x}+\mathbf{\hat y}+\mathbf{\hat z}## is not a unit vector and it has a fixed direction.

The unit vector ##\mathbf{\hat r}## points toward the field point ##(x, y, z)## and therefore has a direction that varies with the field point.
 
  • #11
kuruman
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I agree. I was trying to reconstruct how equation (2) in post #5 was obtained. Sorry about the confusion.
 
  • #12
theshape89
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I see my mistake! Thanks for the correction guys :)
 

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