A) E must be the electric Held due to the enclosed charge
B) If net charge inside the Gaussian surface\[=0,\] then \[\vec{E}\]must be zero everywhere over the Gaussian surface.
C) If the only charge inside the Gaussian surface is an electric dipole, then the integral is zero.
D) E is parallel to \[\overrightarrow{dA}\] everywhere over the Gaussian surface.
Correct Answer: C
Solution :
\[\phi =\frac{{{Q}_{net}}}{{{\in }_{0}}}\]For dipole \[{{Q}_{net}}=0\Rightarrow \phi =0\]
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