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.
Consider a closed surface having certain charges both within and outside as shown. The total electric flux of the given closed surface is : ` (##U_LIK_SP_PHY_XII_C01_E03_011_Q01.png" width="80%">

Options

A. ` + 2.26 xx 10^(5) V m `
B. ` +7.91 xx 10 ^(5) V m`
C. ` -2.26 xx 10 ^(5) V m `
D. ` +1.36 xx 10 ^(-6) V m`

Answer

c

Explanation

`phi_E=(1)/(in_0) sum q_(en) =(-2mu C)/(8.85xx10^(-12) )= )(-2xx10^(-6))/(8.85 xx 10 ^(-12)) =-2.26 xx 10 ^(5) V m ]`