Question

The total electric flux at a point in an electric field is

Answer 1

Hint: We know that electrical flux is defined as the uniform electric field passing through a surface area. The electrical flux can also be regarded as the magnitude of the uniform electric field in the given area of the surface. Let us assume that the surface is closed, and using Gauss law we can solve the question.

Formula used:
 \[{{\Phi }_{E}}=\iint_{s}{E.dS}\]

Complete step-by-step solution: -
Electric flux is the amount or rate of flow of electric field passing through a given area. Clearly, electric flux depends on the number of electrical field lines passing via the area and the angle between the two. According to Gauss Law
\[{{\Phi }_{E}}=\iint_{s}{E.dS}\] where $E$ is the electric field over a small surface area vector$d\;S$
Also, electric flux depends only on the charges which are present inside the area and is symmetrical in nature due to its closed surface property.
Now to find the electric flux at a point in the electric field $E$,clearly, the angle between the electric field and the surface area vector be $\theta$ , as the surface area is a point here, we can say that $dS=0$ ,thus applying gauss law we have
$\phi_e=E(dS) cos \theta=0$
Thus the answer is $0$

Note: The electrical field can either be uniform or non-uniform and the area taken into consideration can either be closed or open surfaces. Gauss law can only be used in closed surfaces, and the flux is generally perpendicular to the surface taken into consideration. Also note that electric flux is a scalar quantity, since $E$ and $d\;S$ are two vectors, the dot product of the two is a scalar.