### Long Question Solution # Electrostatics - Long Question Answer 6

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 = 4πr2 From Gauss theorem, we have
Φ = E A = E 4πr2 .....… (i)
Also from Gauss theorem, we have
Φ = 1/ϵ0 q …..... (ii)

Using equation (i) and (ii), we get
E 4πr2 = 1/ϵ0 q
E = 1/4πϵ0 (q/r2)

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 = 1/4πϵ0 (q/R2)

When point P lies inside the sphere r<R
Since, no charge lies inside the conductor, so q = 0.
Therefore, E = 1/4πϵ0 (0/ R2) = 0

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