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# If the net charge enclosed by a closed Gaussian surface is zero, does this mean that the electric field at all points on the surface is also zero?A) Yes, because $\oint{{\vec{E}}}\cdot d\vec{A}=\dfrac{{{Q}_{enc}}}{{{\varepsilon }_{0}}}$if ${{Q}_{enc}}$is zero, then $\vec{E}$ must be zero.B) Yes; electric field strength depends on how much net charge is present in the region.C) Yes; electric fields can be cancelled out by charges outside the surface.D) No; it is not necessary that the electric field should be zero but flux should be zero.

Last updated date: 12th Aug 2024
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Hint: According Gauss law the surface integral of the electric field intensity over any closed surface (Gaussian surface) in free space is equal to $\dfrac{1}{{{\varepsilon }_{0}}}$times the net charge enclosed by that surface. From this theorem we can see that electric fields and charge enclosed are related. Using this relation we can determine whether the electric field at all points on the surface is zero or not.

Formula used:
$\phi =\oint{{\vec{E}}}\cdot ds=\dfrac{{{q}_{in}}}{{{\varepsilon }_{0}}}$

Consider a surface, with electric field,
Then, flux through the surface is, $\phi =\oint{{\vec{E}}}\cdot d\vec{s}$
Where,
$\vec{E}$ is the electric field
$d\vec{s}$ is the area vector
According to Gauss law,
$\phi =\dfrac{{{q}_{in}}}{{{\varepsilon }_{0}}}$
Then,
$\phi =\oint{{\vec{E}}}\cdot d\vec{s}=\dfrac{{{q}_{in}}}{{{\varepsilon }_{0}}}$ --------- 1
Where,
${{q}_{in}}$ is the charge enclosed
${{\varepsilon }_{0}}$ is the permittivity of free space.
If charge enclosed is zero, i.e., ${{q}_{in}}=0$
$\oint{{\vec{E}}}\cdot d\vec{s}=0$
But it does not imply that the electric field is zero. Dot product of two vectors is zero, when the angle between the vectors is 90 degree, i.e., from the above equation, the electric field can be perpendicular to the surface.
But from equation 1, we can see that, if the charge enclosed by the surface is zero, then total flux through the surface will be zero.
Hence, if the net charge enclosed by a closed Gaussian surface is zero, it is not necessary that the electric field is zero but flux should be zero.

Therefore, the answer is option D.

Note:
The Gaussian surface is a closed surface in three-dimensional space such that the flux of a vector field is calculated. These vector fields can either be the electric field or the gravitational field or the magnetic field. Surfaces like hemisphere, square and the disc are not considered as Gaussian surfaces, because these surfaces do not include 3D volume and have boundaries. Also the Gauss law cannot determine electric field due to electrical dipole.