If $\int_{s} E \cdot d s=0$
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) the number of flux lines entering the surface must be equal to the number of flux lines leaving it
(d) all charges must necessarily be outside the surface
Given,
$\int_{s} E \cdot d s=0$
It means the algebraic sum of number of flux lines entering the surface and number of flux lines leaving the surface.
Now, from Gauss' law,
$\int_{S} E \cdot d s=\frac{q}{\varepsilon_{o}}$
Where q is charge enclosed by the surface.
Now,
$\int_{s} E \cdot d s=0$
q = 0 i.e., net charge enclosed by the surface must be zero.
Hence, all other charges must necessarily be outside the surface.
Question 1.9: The electric field at a point is
(a) always continuous
(b) continuous if there is no charge at that point
(c) discontinuous only if there is a negative charge at that point
(d) discontinuous if there is a charge at that point
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