Question

What will be the value of flux of electric field on a closed surface ABC placed near charges + q and + q 2 and + q 3

Answer 1

Gauss’ law in differential form is as follows:

div D = ρ …(1)

where D is electric displacement, related to electric field E by D = єE (є is electric permittivity) and ρ is charge density. This form of Gauss’ law hold at a point where the charge density is ρ. For a volume distribution of charge in a volume V enclosing a surface S, Eq. (1) is integrated over the volume V:

∫∫∫div D dv =∫∫∫ ρ dv …(2)

Using Gauss’ divergence theorem on the left side of Eq. (2), one gets

∫∫D.nds=∫∫∫ ρ dv = Q …(3)

where Q is total charge enclosed in volume V and integration on th left side s performed on closed surface S. The above relation can be written as

∫∫E.nds=Q/є …(4)

Now left side is electric flux across the surface S. So Eq.( 4) can be stated as “Flux out = Charge within”. In the present case total charge enclosed by the surface is q1+q2-q3. So, the electric flux across the surface is (q1+q2-q3)/є.

Plz mark me brainlist