A square surface of side L m is in the plane of the paper. A uniform electric field \(\rightarrow\) E \(\bigg(\frac{V}{m}\bigg)\), also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is :
- \(\frac{\text{EL}^2}{(2\epsilon_0)}\)
- \(\frac{\text{EL}^2}{2}\)
- Zero
- EL2
The Correct Option is C
Solution and Explanation
Electric flux, E =\(\int\)Ē·\(ds\)
= \(\int\)E\(ds\) cos 0
= \(\int\)E\(ds\) cos 90° = 0 .
Therefore, the correct option is (C): Zero
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Concepts Used:
Gauss Law
Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.
Gauss Law:
According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.
For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.
Gauss Law Formula:
As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;
Q = ϕ ϵ0
The Gauss law formula is expressed by;
ϕ = Q/ϵ0
Where,
Q = total charge within the given surface,
ε0 = the electric constant.