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In a question of my book it says that an electric quadrupole is a system of two dipoles of equal magnitude but opposite in sign. If that's the case and since they are vectors, won't they cancel out and cause the net dipole to be zero? Apart from this can someone tell me how to find out the net electric field on a point on the dipole axis of an electric quadrupole?

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    $\begingroup$ Related question by OP: physics.stackexchange.com/q/322960/2451 $\endgroup$ – Qmechanic Apr 2 '17 at 6:24
  • $\begingroup$ that is not about quadrupole. quadrupole is made up of two dipoles $\endgroup$ – Esha Mukhopadhyay Apr 2 '17 at 6:25
  • $\begingroup$ Qmechanic just pointed out your other related question, which you take here to the next step but without any reaction to the previous one. Imho you could find the answers with a few minutes of basic physics reading. $\endgroup$ – Helen - down with PCorrectness Apr 2 '17 at 6:31
  • $\begingroup$ yea but y dont the dipoles cancel each other? $\endgroup$ – Esha Mukhopadhyay Apr 2 '17 at 7:22
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    $\begingroup$ The total charge gets canceled, but each point in space sees a different asymmetrical distribution of charges around it. Also, @ZeroTheHero's link. $\endgroup$ – Helen - down with PCorrectness Apr 2 '17 at 19:07
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Quadrupoles arise from dipoles in the same way that dipoles arise from point charges: you take two opposing ones at a finite separation $\mathbf s$, and then you take the limit of $s\to0$ while the dipoles' moments $p$ grow to infinity in such a way as to keep $ps$ constant. And, in the same way that the point charges' fields do not fully cancel out (giving us dipolar fields that decay as $1/r^3$ instead of $1/r^2$), the dipoles' fields don't fully cancel out either, yielding a zero total dipole but producing a quadrupolar field that decays as $1/r^4$. It's all exactly analogous to the monopole-to-dipole limit.

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