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The electric field of the electric dipole is not zero. Since the charge $q$ and $–q$ are separated by some distance, the electric fields due to them, when added, do not exactly cancel out. However, why do the fields due to $q$ and $–q$ cancel out at distances much larger than the separation of the two charges forming a dipole ($r \gg 2a$)?

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    $\begingroup$ Who says that they cancel out? $\endgroup$ – Emilio Pisanty Apr 2 '17 at 13:39
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    $\begingroup$ They don't cancel out. -1 en.wikipedia.org/wiki/… $\endgroup$ – Rob Jeffries Apr 2 '17 at 14:52
  • $\begingroup$ They don't cancel, a simple vector analysis of the fields will give you the clue. Infact at large distance the field falls as $\frac1{r^3}$. $\endgroup$ – sbp Apr 2 '17 at 15:57
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The total electric field is a sum of two fields: $${\bf{E}}({\bf{r}}_1)+{\bf{E}}({\bf{r}}_2)\propto q\left(\frac{1}{r_1^2}-\frac{1}{r_2^2}\right)=q\frac{\left(r_2^2-r_1^2\right)}{r_1^2r_2^2}\approx q\frac{a}{r^3},\;\;r_1\approx r_2=r$$ At large distances the denominator is much larger than the numerator; that is why the total field is much less than that from one charge $\propto 1/r^2$ by the factor $a/r\ll 1$.

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  • $\begingroup$ Sir, please explain me why is it written that the fields cancel out? $\endgroup$ – Adheeti Agarwal Apr 2 '17 at 12:15
  • $\begingroup$ The electric field of one charge subtracts the electric field of the other one. At large distances this subtraction is nearly perfect, thus the word "cancellation" makes sense. $\endgroup$ – Vladimir Kalitvianski Apr 2 '17 at 13:20
  • $\begingroup$ The use of the word cancel here is imprecise and not strictly correct. The fields tend to cancel the farther out you go, but they never completely cancel. At some point the field will be so close to zero that it will be unmeasurable by any apparatus that you can imagine, at which place the field will be zero for all practical purposes. $\endgroup$ – garyp Apr 2 '17 at 13:48
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The whole point of electric dipoles is that the fields from the two charges don't cancel out. The importance of the limit where the distance $r$ from the dipole's center is much bigger than the intercharge separation $2a$, i.e. $r\gg 2a$, is that the internal details of the field don't matter as much, but you're not left with zero; instead, you're left with a residual field of the form \begin{align} \phi(\mathbf r) & = \lim_{r\gg2a}\left[\frac{q}{\|\mathbf r-a\hat{\mathbf u}\|}-\frac{q}{\|\mathbf r+a\hat{\mathbf u}\|}\right] \\ & = \frac{p \hat{\mathbf u}\cdot \hat{\mathbf r}}{r^2} = \frac{\mathbf p\cdot\mathbf r}{r^3} \end{align} which decays as $1/r^2$, i.e. faster than the $1/r$ decay of the Coulomb potential, but which is nevertheless not zero. This is because the fields generally act against each other, but because they act with different strengths and in different directions, it is impossible for them to fully cancel out.

If your text states that the fields actually cancel out, then your text is likely to be wrong. However, unless you provide a reference or the passage in question, it is impossible to tell whether the text is actually wrong or you're just misreading its contents.

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