IIT JEE , AIEEE Forum - Physics , Electricity

edison

Dear pinnaka,

This is to explain your query sent to me in my nudge book.

You said that i have taken Electric field to be constant on the bottom of the cylinder.

But this is not the case, I have not take field constant.

Rather I tried to find the surface integral of electric field by integrating over the surface.

And for the sake of this I considered a ring of inner radius r and outer radius r + dr on the circular base/ bottom of the cylinder.

And ofcourse the Electric field on this ring can be considered as constant which is taken as

E = k Q / (L²/4 + r²)

Now the this is integrated over the surface of circular base.

Remember here radius of ring = r

But radius of circular base = R

and 0 < = r < = R

Hope your doubt is now cleared.