Question

Electric field at centre of uniformly charged hemisphere

Answer 1

Answer:

We take a surface element da in the direction ( theta), (phi).

With origin of spherical polar coordinate system at the center o of the hemisphere, we have,

da=R^2 [d( omega)]=R^2 sin(theta) d(theta) d(phi). Here d(omega) is solid angle element).

Charge element on this surface element,

dq==(6)da=6 R^2 sin(theta) d(theta) d( phi)…………(1)

Note: read 6 as sigma.

The field at o ( the center) , due to this charge element is

dE=(kdq/R^2)…………….(2)

Take two components of dE. One parallel to base. This is dE cos (theta). When this component is integrated, it's contribution is zero.

The component dE sin (theta) is of interest.

Now, dE sin(theta)=(k dq/R^2) sin (theta) , from eq.(2).

Taking dq from eq.(1),

dE sin(theta)=(k/R^2) sin (theta)(6)R^2 sin(theta) d(theta) d(phi)

Integrating dE sin(theta) over ( theta) from 0 to pi and over ( phi) from o to 2 pi,we see that d(phi) will be integrated to 2(pi). Then there will remain the following integral over ( theta). Thus,

Field at center, E=(2 pi k 6) integral from 0 to pi [ sin^2(theta) d( theta)……….(3)

Using the fact that integration of sin^2( theta) is[(theta)/2 - (sin2(theta)/4] and putting limits (theta) from 0 to pi, we get

E=2 pi k 6 pi/2=pi k 6=pi 6/(4 pi eo)=6/(4 eo).

eo is permittivity of space.