**Question: Use Gauss theorem to calculate the electric field due to a solid charged sphere at a point inside it and outside the sphere.**

**Ans:**Let us consider a charged sphere of radius R has charge q on its surface. There is a point P at a distance r from the center O of the charged sphere, where we have to find the electric field (intensity).

We now draw a concentric sphere of radius r so that the point P will lie outside from the surface of the sphere. Therefore, Area of this sphere = 4πr

^{2}^{}From Gauss theorem, we have

Φ = E A = E 4πr

^{2}.....… (i)Also from Gauss theorem, we have

Φ = 1/ϵ

_{0}q …..... (ii)Using equation (i) and (ii), we get

E 4πr

^{2}= 1/ϵ_{0}qE = 1/4πϵ

_{0}(q/r^{2})Thus, the electric field outside the charged sphere is same as while the charge were concentrated at its center.

When the point p lies at the surface of the charged sphere (R = r). electric field intensity at this surface is

E = 1/4πϵ

_{0}(q/R^{2})When point P lies inside the sphere r<R

Since, no charge lies inside the conductor, so q = 0.

Therefore, E = 1/4πϵ

_{0}(0/ R^{2}) = 0
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