Dipole field and ideal dipole

A dipole field is the electric field produced by an electric dipole. The net charge on an electric dipole is zero but it has an electric field because both the charges are placed at some distance from each other. So, there is always an electric field due to an electric dipole.

But in the case of an ideal dipole, if charge $Q$ gets larger and separation distance get smaller and smaller by keeping the product $Q\times\mathrm{distance} = \mathrm{constant}$, then the dipole is said to be an ideal dipole.

So, when separation distance starts getting really close to the zero, what happens to the dipole field? Does it also go to zero and vanish?

• No, because the increase in $Q$ compensates for the decrease in separation! You need to look at the maths! – Philip Wood Jun 22 '17 at 7:58
• @PhilipWood But the only reason why the dipole had an electric field was because it's charges were placed at some distance from each other. When the separation distance approaches zero then the dipole field should also approach zero? – M. Singh Jun 22 '17 at 9:18
• – Emilio Pisanty Jun 22 '17 at 11:13
• @M.Singh 'Approaches' is the key word! Have you done the maths? – Philip Wood Jun 23 '17 at 10:29
• @PhilipWood What are you proposing? – M. Singh Jun 23 '17 at 14:45

If you have a finite dipole, i.e. a pair of charges a finite distance $d$ apart, with a finite charge $q$, then the field will not be purely dipolar: the leading contribution at distances $r\gg d$ will be dipolar, going down with $r$ as $1/r^3$, but there will also be other contributions to the field: the subleading term is an octupolar field, which goes down as $1/r^5$, and it goes down from there. (Quadrupolar and hexadecapolar fields do not contribute because of symmetry restrictions.)
The point of taking the limit from the finite dipole to a point dipole (also called an ideal dipole), by making $d\to0$ while keeping $p=qd$ constant, is to retain this dipole field while boiling away the higher-order contributions.
It's important to note how crucial it is to have the charge go to infinity at the precisely controlled rate of $q=p/d\to\infty$. If you keep the charge fixed, and then you take the limit $d\to0$, then obviously the field will vanish in the limit, because it keeps getting smaller and smaller as $d$ decreases.
In the point-dipole limit, however, when we say $d\to 0$, at each step of the limit $d$ is still positive, so you have a nonzero separation between the charges, and therefore a nonzero field, which is (roughly) proportional to the product $p=qd$. Then, if you make the charge increase at the correct rate, you boost the field (which would go to zero at constant $q$) back up to a constant level, in a way that will give it a definite nonzero limit as $d\to0$.
Thus, in the $d\to0$ limit, the field will be a pure dipole field, $$\mathbf E(\mathbf r) = {\frac {1}{4\pi \epsilon _{0}}}\frac {3(\mathbf {p} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {p} }{r^{3}},$$ at all positions (not just in the $r\gg d$ far field), and it will not vanish.