The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by
- $ 4 \,\pi\varepsilon_0\, Aa^3 $
- $ \varepsilon_0\, Aa^3 $
- $ 4 \, \pi\varepsilon_0 \, Aa^2 $
- $ A \, \varepsilon_0 \,a^2 $
The Correct Option is A
Solution and Explanation
E = Ar
Here r is the radial distance.
At r = a, E = Aa ...(i)
Net flux emitted from a spherical surface of radius a is $ \phi_{net} = \frac{q_{en}}{\varepsilon _0} $
$ \Rightarrow (Aa) \times (4\pi a^2) = \frac{q}{\varepsilon _0}$ [Using equation (i)]
$ \therefore q = 4\pi\varepsilon _0Aa^3 $
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Concepts Used:
Gauss Law
Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.
Gauss Law:
According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.
For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.
Gauss Law Formula:
As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;
Q = ϕ ϵ0
The Gauss law formula is expressed by;
ϕ = Q/ϵ0
Where,
Q = total charge within the given surface,
ε0 = the electric constant.