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

Options

A. `A in_(0) a^(2)`
B. `4 pi in_(0) A a^(3)`
C. `in_(0) A a^(3)`
D. `4 pi in_(0) A a^(2)`

Answer

B

Explanation

Flux linked with sphere `= vec(E). D. vec(s)`.
Since electric field is radial.It is always perpendicualr to the surface of the sphere.
So `phi = EA cos 0^(@) = EA = Ar (4 pi r^(2))`
`phi = A pi a ^(3)` (as `r = a`)
According to gauss's law,
`phi = (q)/(in_(0))` or `q = in_(0) phi`
`q = in_(0) A (4pi a^(3)) = 4pi in_(0) A a^(3)`