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If $\int_s \mathrm{E} . \mathrm{dS}=0$ over a surface, then
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The correct answers are:
the number of flux lines entering the surface must be equal to the number of flux lines leaving it
,
all charges must necessarily be outside the surface
the number of flux lines entering the surface must be equal to the number of flux lines leaving it
,
all charges must necessarily be outside the surface
In general Gauss's law is applied to any closed surface.
In Gauss's law, $\oint$ E.dS means the algebraic sum of number of flux lines entering the surface and number of flux lines leaving the surface.
When $\oint_{\mathrm{s}} E \cdot \mathrm{dS}=0$, the number of flux lines entering the surface must be equal to the number of flux lines leaving it. If there is no change inside.
$\oint_{\mathrm{s}} \mathrm{E} . \mathrm{dS}=\frac{\mathrm{q}}{\varepsilon_0}$, where $\mathrm{q}$ is charge enclosed by the surface, and $\epsilon_0$ is permitivity.
If $\oint \mathrm{E} \mathrm{dS}=0, \mathrm{q}=0$ then the resultant charge enclosed by the surface must be zero.
So, all other charges must necessarily be outside the surface. This is because charges outside because of the fact that charges outside the surface do not contribute to the electric flux.
In Gauss's law, $\oint$ E.dS means the algebraic sum of number of flux lines entering the surface and number of flux lines leaving the surface.
When $\oint_{\mathrm{s}} E \cdot \mathrm{dS}=0$, the number of flux lines entering the surface must be equal to the number of flux lines leaving it. If there is no change inside.
$\oint_{\mathrm{s}} \mathrm{E} . \mathrm{dS}=\frac{\mathrm{q}}{\varepsilon_0}$, where $\mathrm{q}$ is charge enclosed by the surface, and $\epsilon_0$ is permitivity.
If $\oint \mathrm{E} \mathrm{dS}=0, \mathrm{q}=0$ then the resultant charge enclosed by the surface must be zero.
So, all other charges must necessarily be outside the surface. This is because charges outside because of the fact that charges outside the surface do not contribute to the electric flux.
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