IIT JEE , AIEEE Forum - Physics , Electricity

priyeshpatel

IF THE RADIUS OF THE GUASSIAN SURFACE ENCLOSING CHARGE IS HALVED, HOW DOES THE FLUX THROUGH THE GUASSIAN SURFACE CHANGE ..........

neeraj_agarwal_1990

flux depends only on the charge.... its q/e (e is epsilon)

so it will remain the same...

bsgdabest

@ neeraj:
i dont think that it is necessary to remain the same, consider a solid sphere of rad r charged over total volume, and gaussian surface has a radius r/2?( it is not given as "point charge" rite??)

neeraj_agarwal_1990

but i think the charge mentioned should be a point charge....dats why is the question made!

the question cannot be answered any other way if its objective...

harsha_27

Priyesh......Your question is incomplete.
If it is a point charge then Neeraj is correct.
Otherwise it must be mentioned abt the distribution of charge.

edison

If radius of gaussian surface is halved and charge still remains within the gaussian surface, then FLUX THOUGH THE SURFACE REMAINS THE SAME

As, Flux = q/

₀

So it does not change as it is independent of area

But the parameter that changes here is FLUX DENSITY i.e. Flux/Area

So initally area is = A = 4

r²

then if radius is halved A' = 4

r²/4 =

r²

So A/A' = 4

Hence, flux density increases by a factor of 4

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