Answer questions on the basis of your understanding of the following paragraph and the related studied concept.
Electric flux, in general , through any surface is defined as per relation:
`phi_E= int oversetto E. oversetto (ds) ` , where integration has to be performed over the entire surface through which flux is required. The surface under consideration may be a closed one or an open surface. When flux through a closed surface is required we use a small circular sign on the integration symbol. Thus flux over a closed surface `oint E= oint oversetto E. oversetto (ds) . ` it is customary to take the outward normal as positive in this case.
A German physicist Gauss established a fundamental law to find electric flux over a closed surface. As per Gauss' law , the flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by `in_0 . `Mathematically `oint oversetto (E) . oversetto (ds) =(1)/( in_0) [q_(en) ] ` , where `q_(cn) ` is the net charge enclosed within the surface. It is possible to derive Gauss' law from Coulomb's laws. Gauss' law can be applied to obtain electric field at a point due to continuous charge distribution for a number of symmetric charge configurations.
A square surface of side L m is in the plane of paper. A uniform electric field ` oversetto E V m ^(-1) ` , also in the plane of the paper, is limited only to the lower half of the square surface as shown in adjoining figure. The electric flux, in SI units, associated with the surfer is :
` (##U_LIK_SP_PHY_XII_C01_E03_012_Q01.png" width="80%">

Options

A. zero
B. `EL^(2) `
C. ` (EL^(2))/( 2 in _0) `
D. ` (EL^(2))/(2) `

Answer

a

Explanation

Since electric field ` oversetto E ` lies parallel to the surface area , `oversetto E and overesetto S ` are mutually perpendicular to each other and hence net electric flux will be zero.