1.

The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a shepere of radius 'a' centred at the origin of the field, will be given by :

Answer»

`Aepsilon_(0)a^(2)`
`4piepsilon_(0)Aa^(3)`
`epsilon_(0)Aa^(3)`
`4piepsilon_(0)Aa^(2)`

Solution :Flux linked with SPHERE =`vecE.dvecS`
since electric FIELD is RADIAL. It is always perpendicular to the surface.
So, `phi =Ar.(4pir^(2))`
`phi = A(a)(4pir^(2))`
`phi =A4pia^(3)` (as r=a)
Now according to gauss law
`phi = (q_(in))/epsilon_(0) RARR q_(in) =phi.epsilon_(0)`
So,= `q_(in) =Apia^(3)epsilon_(0)`


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