Electric Field due to Uniformly Charged Shpere

A spherically symmetric charge distribution means the distribution of charge where the charge density p at any point depends only on the distance of the point from the centre and not on the direction. Consider a total charge q distributed uniformly throughout a sphere of radius R. Note that the sphere cannot be a conductor or, as we have seen, the excess charge will reside on its surface.

Case (i). When the point lies outside the sphere

P is a point at a distance r from the centre O. We have to find the electric field E at P. Draw a concentric sphere of radius OP with centre O. This is the Gaussian surface. At all points of this sphere, the magnitude of the electric field is the same and its direction is perpendicular to the surface. Angle between E and S is zero.
The flux through this surface is given by
Ф E. dS = Ф E dS = E (4π r2)

By Gauss’s law
E (4π r2) = q/ε₀ or E = 1/ (4π r²) q/ r²

In vector form
E = 1/ (4π r²⁾ q/ r² r

Hence the electric field at an external point due to a uniformly, charged sphere is the same as if the total charge is concentrated at its centre.

Case (ii). When the point lies on the surface.

Here, r = R
E = 1/ (4π r²) q/ R²

Case (iii). When the point lies inside the sphere

P’ is a point inside the sphere. P’ is at a distance r from the centre O. Draw a concentric sphere of radius r (r<R) with centre at 0. This is the Gaussian surface.

Total; charge enclosed by the Gaussian surface   
q’ = 4/3 π r³ p   
    = 4/3 π r³ ___q___ = q r³
          ______4/3 π R³       R³

Here, p = charge density = char geper unit volume = ___q___
                             ____________________________ 4/3 π R³

The outward flux through the surface of the sphere of radius r is
Ф E . dS = E (4π r²)

Applying Gauss’ law,

E (4π r²) = q’/ ε₀ = q/ε₀ = q/ε₀ r³ / R³
. :  E = ___1____ qr
    ______4π ε₀    R³

Thus E ∞ r. At the centre of the sphere, E = O.