Two parallel infinite opposite charged lines (of uniform density $\lambda$) are placed at distance $L$ from each other. Need to calculate the electric dipole moment of this system.

How do you do such a calculation? do I need to sum the moment dipoles of infinite tiny charges?

  • $\begingroup$ Do you mean the electric dipole moment? Are the two lines oppositely charged? $\endgroup$
    – Karthik
    Apr 12 '19 at 16:39
  • $\begingroup$ yes, sorry I didn't mention $\endgroup$
    – Frogfire
    Apr 12 '19 at 16:43

The dipole moment of an arbitrary charge density distribution $\rho(\mathbf r)$ is defined as $$ \mathbf d = \int \mathbf r \, \rho(\mathbf r) \:\!\mathrm d\mathbf r, $$ where $\rho(\mathbf r) \:\!\mathrm d\mathbf r$ needs to be understood as a singular measure (say, using suitable delta-function distributions) in the case of point, surface or line charges. For the example of a line charge parametrized by $\mathbf r(s)$ over some parameter $s\in (a,b)$ and with longitudinal charge density $\lambda(s)$, this reads $$ \mathbf d = \int_a^b \mathbf r(s) \:\! \lambda(s) \:\!\mathrm ds. $$ If you have two infinite straight line charges at separation $a$ with constant longitudinal charge density $\lambda$ and with orthogonal unit separation vector $\mathbf u$, then this calculation yields a constant dipole moment $\Delta \mathbf p = \lambda \:\!\Delta L \:\! a\mathbf u$ per stretch of length $\Delta L$ of the two wires. The total dipole moment of the system can be inferred from that.

  • $\begingroup$ For "infinite" wire like the OP said, that would be like infinite dipole moment is it? $\endgroup$
    – Karthik
    Apr 12 '19 at 17:00
  • $\begingroup$ but Δp=λΔLau means p diverges for infinite wires, isn't it? $\endgroup$
    – Frogfire
    Apr 12 '19 at 17:02
  • $\begingroup$ @Frogfire This answer has some intentional gaps that are there for you to figure out yourself. I hope this was clear from the get-go. If you find that conclusion shocking, then what expectations did you have that led you to expect a different result? $\endgroup$ Apr 12 '19 at 17:07
  • $\begingroup$ @EmilioPisanty Well getting infinities always used to mean in our old literature as to something has gone wrong! :) But ya, I am convinced of this answer anyways. $\endgroup$
    – Karthik
    Apr 12 '19 at 17:21
  • 1
    $\begingroup$ @KV18 I'm not sure what all the fuss is about. If you start with a plainly unphysical situation, why are people surprised that the results are unphysical? $\endgroup$ Apr 12 '19 at 17:26

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