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If we assume to place a dipole or a quadrupole inside a Gaussian surface, then the Gauss law will give zero electric field outside Gaussian surface at any point since enclosed change is zero. However, we know that electric field will surely be there because, say, in case of dipole, no neutral point where electric field vanishes will exist.

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Gauss’s Law is about the integral over the entire surface. Consider a charge-free Gaussian volume in a uniform external electric field. The flux integral

$$ \Phi = \int \vec E \cdot \mathrm d\vec S $$

vanishes, not because $\vec E$ is zero anywhere, but because every electric field line that enters the surface leaves it again.

For the dipole field,

$$ \vec E_\text{dipole} \propto \frac{3(\vec p \cdot \hat r) \hat r - \vec p}{r^3} $$

consider integrating over a sphere with radius $R$ centered on the dipole. The surface element of the sphere is

$$ \mathrm d\vec S = \hat r\ R^2\ \mathrm d\Omega = \hat r\ R^2\ \mathrm d(\cos\theta)\ \mathrm d\phi $$

so both terms in the field are reduced to $\vec p \cdot \hat r = |\vec p| \cos\theta$. That gives a total flux

$$ \Phi_\text{dipole} \propto \int_{\cos\theta=-1}^{+1} \cos\theta\ \mathrm d\Omega = 0 $$

The flux integral vanishes, even though the flux vanishes only on the “equator” of the sphere, and even though the field is nonzero everywhere.

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If we assume to place a dipole or a quadrapole inside a Gaussian surface then gauss law will give zero electric field outside Gaussian surface at any point since enclosed change is zero

Gauss' law does not say the electric field is zero outside the surface if the net enclosed charge is zero. It says that the net electric flux across the surface is zero if the net enclosed charge is zero.

..but we know that electric field will surely be there because say in case of dipole no neutral point where electric field vanishes will exist

Yes, but there is no violation of Gauss' law for the reason already stated.

Hope this helps.

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You could use gauss law in this case to find de flux of electric field through a random surface that surrounds the dipole, as every field line that leaves the dipole will come back and its contribution to the flux cancels.

What you can't do in this case is find the electric field with gauss law directly as you need some simetry to get E out of the integral.

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