Question

drive expression for electric field due to uniformly charged thin spherical shell​

Answer 1

Answer:

The electric field of a sphere of uniform charge density and total charge charge Q can be obtained by applying Gauss' law. Considering a Gaussian surface in the form of a sphere at radius r > R, the electric field has the same magnitude at every point of the surface and is directed outward. The electric flux is then just the electric field times the area of the spherical surface.

pi= EA

=E4πr^2

=Q/E•

The electric field outside the sphere (r > R)is seen to be identical to that of a point charge Q at the center of the sphere.

For a radius r < R, a Gaussian surface will enclose less than the total charge and the electric field will be less. Inside the sphere of charge, the field is given by:

Answer 2

Electric field due to uniform charge spherical cell

Solution

Assume that the radius of cylinder be R

the sphere is surrounded by Gaussian surface which is spherical in shaphrical in shape having direction is perpendicular to its surface .

Assume that the direction of electric field is parallel to Gaussian surface direction.

Formula

Area of sphere = 4πR^2

we know that from gauss law

d(electric flux ) = E.ds ------(1

Electric flux is dot product of electric field and gaussian surface .

i assume small area ds and with in small flux

integrate equation ------(1 on both side

so

Electric flux = E [4πR^2]

Electric field = Electric flux/4πR^2-------------(2)

Again we know that

Electric flux = q(in)/abselent node(Eo) -------------(3)

Here q(in) is the charge inside the spherical gaussian surface .

the charge present in sphere is on its outside surface since charge is not present inside the

sphere .

the abselent node is surrounding medium of gaussia

surface

on combining equation -----(2 and -------(3 we get

Eletric field = qin/4πEoR^2