# Application of gauss law for electrostatics: Electric Field Due To A Uniform Charged Sphere

Let us today discuss another application of Gauss law for electrostatics that is the Electric Field Due To A Uniform Charged Sphere:-

Consider a charge +q be uniformly distributed in a sphere of radius R with centre at O.

The charge density p at every point is uniform.

Let P be the point at a distance r from the centre of the charged distribution at which the intensity of electric field is to be determined (try to make the figure yourself as assignment).

Here, three cases arises:-

(1) Electric Field Due To A Uniform Charged Sphere At An External Point(r>R):-

Let point P lies outside the charged sphere , In order to find the intensity of electric field at P,we draw a concentric Gaussian spherical surface of radius r through P. All points on this surface are at the same distance from the centre. Due to spherically symmetric charge distribution the magnitude of intensity of electric field at all points on the Gaussian sphere is same. As q is positive, the field is radially outward in direction.Therefore,the electric flux dφ through any area element dS taken on the surface,where E and area vector dS both are directed radially outward ,that is ,the angle between them is zero ,is given by

Dφ=E.dS=Edscos0^{0}=EdS

Thus, the electric flux through the entire Gaussian sphere as discussed in the gauss law for electrostatics is

Φ=∫E.dS=∫EdS

Or φ=E∫dS

But ∫dS=4πr^{2}

Φ=E(4πr^{2}) (1)

According to the Gauss’s law,

Φ=q/ε_{0} (2)

By comparing equations(1) and (2),we get

E(4πr^{2})=q/ε_{0}

Or E=q/4πε_{0}r^{2} (3)

Conclusion .The elementary of electric field due to a uniformly charged sphere at an external point (r>R) is identical with the field intensity due to an equal point charge placed at the centre or charged sphere behaved as if its entire charge is concebtrated as its centre.

(2) Electric Field Due To A Uniform Charged Sphere At a Point on the surface of the sphere (r=R):-

If the point P is situated t the surface of uniformly charged sphere,then the distance of P from the centre of the sphere O is equal to its radius ,that is r=R

Thus, the intensity of electric field at the surface of the sphere is obtained by replacing r by R in equation(3)

Thus E=1/4πε_{0} q/r^{2}

(3) Electric Field Due To A Uniform Charged Sphere At an Internal Point(r<R):-

In order to determine the intensity of electric field at a point P inside the charged sphere, distant r from the centre O,we draw a concentric Gaussian spherical surface of radius r through P, For any small area element dS taken on the surface ,area vector dS and field vector E both are in the same direction.

Therefore, the electric flux φ through the entire Gaussian surface is

Φ=∫E.dS=∫EdS=E(4πr^{2}) (4)

Let us now calculate the total charge contained inside the Gaussian sphere. As the charge is uniformly distributed throughout the uniformly charged sphere, volume charge density ρ is

Ρ=total charge/volume=q/(4/3)R^{3}

The volume occupied by the Gaussian sphere =4/3 πr^{3}

The total charge contained by the Gaussian sphere

Q_{1}=ρ*volume

Or q_{1}=[q/(4/3πR^{3}].4/3πr^{3}

Or q_{1}=qr^{3}/R^{3} (5)

According to Gauss’s law

Φ=∫E.dS=q_{1}/ε_{0} (6)

Comparing equations (4) and (6),we get

E(4πr^{2})=q_{1}/ε_{0} (7)

Substituting the value of q_{1} from equation (5) in equation (7),we get

E(4πr^{2})=1/ε_{0}(qr^{3}/R^{3})

Or E=qr/4πε_{0}R^{3}

Conclusion: The intensity of electric field due to a uniformly charged non-conducting sphere at an internal point is directly proportional to the distance of the point from the centre,