Question

Obtain an expression for the electric intensity at a point outside a uniformly charged spherical shell.

Answer 1

Consider an isolated charged hollow spherical conductor A with radius R and surface charge density σ, placed in a medium with permittivity ε. Consider a point P outside the conductor, r distance from its centre. To determine the electric field intensity at P, we select a spherical Gaussian surface S with radius r that passes through P and is concentric with conductor A. A little portion of this surface containing P has an area dS.

The charge on the sphere is:

Q = σ(4πr²)

The charge Q is equally distributed on the outer surface of the spherical conductor. Then, by symmetry, the electric field intensity at each point on surface S is normal to the surface and has the same magnitude E. If charge Q is positive, the `vecE` at each point on S is radially outwards.

When the angle θ between `vecE` and `dvecS` is zero for all surface elements, the electric flux through them equals.

`dphi = vecE*dvecS`

= E dS

Therefore, the flux through the Gaussian surface S is:

`phi = oint E dS`

= `E oint dS`

`oint dS` = surface area of the sphere = 4πr²

∴ `phi = E xx 4pir^2`

Then, by Gauss’s theorem,

`phi = Q/ε = E xx 4pir^2`

E = `Q/(4piεr^2)`

= `Q/(4piε_0kr^2)`