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Question: If \(\oint_{}^{}{s\overrightarrow{E}.\overrightarrow{d}s = 0}\)over a surface, then...

If ∮sE→.d→s=0\oint_{}^{}{s\overrightarrow{E}.\overrightarrow{d}s = 0}over a surface, then

A

The electric field inside the surface and on it is zero

B

The electric field inside the surface is necessarily uniform.

C

All charges must necessarily be outside the surface.

D

All of these.

Answer

All charges must necessarily be outside the surface.

Explanation

Solution

: As ∮E→.ds→=qε0\oint_{}^{}{\overrightarrow{E}.d\overrightarrow{s} = \frac{q}{\varepsilon_{0}}}

Where q is charge enclosed by the surface.

When ∮E→.ds→=0,q=0\oint_{}^{}{\overrightarrow{E}.d\overrightarrow{s} = 0,q = 0}

i.e., net charge enclosed by the surface must be zero. Therefore, all other charges must be outside the surface. This is because charges outside the surface do not contribute to the electric flux.