Write the expression for electric field intensity at any point outside and inside due to a charged spherical shell.

Explanation

Gauss.s law : The total electric flux through closed a surface in free space is equal to `1//epsilon_(0)` times the net enclosed by the surface.
Derivation : Consider a thin spherical shell of radius R with center O. Let a charge +q is uniformly distribution over the surface of the shell. Let P be a point at a distance r from O. It is desired to find electric intensity at P. Imagine a sphere (dotted) with center O and radius r. The surface of this sphere is Gaussian surface. The electric intensity E is same at all points of Gaussian surface and directed radially outward. The Gaussian sphere is assumed to be divided into small area elements each of area `Delta S`.

The electric flux through the Gaussian surface is given by,
`phi = Sigma E cos theta.Delta S` ......(1)
The angle `theta` between `vec(E)` and `Delta vec(S)` is ` therefore cos 0 = 1`
`(1) rArr phi = Sifma E Delta S`
`phi = E Sigma Delta S` ......(2)
Where `Sigma Delta S` = area of the Gaussian surface `= 4pi r^(2)` [area of the sphere]
`(2) rArr phi = E xx 4pi r^(2)` ....(3)
From Gauss Theorem, `phi = (q)/(epsilon_(0))` .......(4)
On comparing the equation (3) & (4), we get
`E xx 4pi r^(2) = (q)/(epsilon_(0))`
`E = (q)/(4pi epsilon_(0)r^(2))`
`rArr E = (1)/(4pi epsilon_(0)r^(2))`