# Application of Gauss law of electrostatics: Electric Field Due To Two Thin Concentric Spherical Shells

Let us again discuss another application of Gauss law of electrostatics that is Electric Field Due To Two Thin Concentric Spherical Shells:-

Consider charges +q_{1} and +q_{2} uniformly distributed over the surfaces of two thin concentric metallic spherical shells of radii R_{1} and R_{2} respectively

In order to determine the electric field E at a point P, distant r from the centre O by using Gauss’s law ,draw a concentric Gaussian spherical surface of radius r through P. Due to symmetric distribution of charge,the mgnitude of E at all points on the Gaussian surface will be same and radially outward in direction (Try to make figure yourself).

Thus, for any area element dS taken on the gaussian surface,the field vector E and area vector dS both are parallel, therefore,

E.dS=EdS cos 0^{0}=EdS

Hence, the flux through the entire Gaussian surface will be

Φ=∫E.dS=∫EdS=E∫dS

But ∫dS=4πr^{2}

Φ=E(4πr^{2}) (1)

According to Gauss’s law for electrostatics

Φ=q/ε_{0 }(2)

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

E(4πr^{2})=q/ε_{0}

Or E=q/4πε_{0}r^{2} (3)

Three cases arises here:-

(i) Electric Field at a Point Inside the inner Shell of Two Thin Concentric Spherical Shells(r<R_{1}):-

In this case there would be no charge within the Gaussian surface.Therefore,according to Gauss ‘s law q=0.Thus,E=0 at a point inside the inner shell.

(ii) Electric Field at a Point Between the two Shells of Two Thin Concentric Spherical Shells(R_{1}<r<R_{2}):-

In this case,consider that the point P at which the electric field is to be determined lies between R_{1 }and R_{2}.Therefore, the net charge enclosed by the Gaussian sphere is only the charge residing on inner shell,that is q_{1} alone. Therefore, from equation (3)

E= q_{1}/4πε_{0}r^{2}

(iii) Electric Field at a point outside the outer shell of Two Thin Concentric Spherical Shells:-

In this situation ,consider that the point P lies outside the outer shell at a distance r from the centre O,such that r>R_{2}.In this case the net charges enclosed by the Gaussian surface will be the sum of charges on both shells,that is,

Q=q_{1}+q_{2}

Thus from equation(3),

E=q_{1}+q_{2}/4πε_{0}r^{2}

If two spherical shells have charges equal and opposite that is, if one shell has +q charge and the other have =q .In this situaion,inside the inner spherical shell and outside the outer shell the electric field is zero.

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