Understanding the Electric Field of a Hemispherical Capacitor

[cragar]

I saw this setup in a book. Two hemispheres separated by a small distance, one with -Q charge and one with +Q charge. The hemispheres are setup as if I took a sphere and cut it in half and moved one up slightly, and the one on top has the +Q charge. The direction Of the E field right in the middle of the 2 hemispheres should be pointing straight down right? The E field points from plus to minus. I was thinking of it as a hemispherical capacitor. And also I was wondering if the E field outside the 2 hemispheres would be zero because the charge enclosed would be zero, but its not evenly distributed. This is not home work. Any help will be appreciated.

 

[Ken G]

The field outside won't be zero, because Gauss' law tells you the integrated field is zero, but you don't have a symmetry that let's you know the field is zero at each location.

 

[cragar]

ok thanks , could you integrate over each hemisphere and add them together to get each field.

 

[Ken G]

Yes, I think this problem calls for an integral. Are you given that the charges are evenly distributed over the hemispheres? The integral is simplified if you only want the field at the central point, because then all the contributions have the same magnitude but have a different component of that magnitude that points in the vertical direction. The magnitude you can get from Gauss' law (just put Q at the central point and find its field at the radius of the sphere, then note the situation you actually want has that same total field strength except reduced by the fact that you are just taking the vertical component).

The integral you have to do is an integral over the fraction of the field that is vertical, as you vary over the surface area of the sphere. The surface integral looks like sin(theta) dtheta, because the phi coordinate is symmetric, and the vertical component looks like cos(theta). Often you replace cos(theta) by mu, and integrate mu from 0 to 1, and the integral you are doing is then mu*dmu. So all you are doing is an integral from 0 to 1 of mu*dmu, and comparing it to the field you'd get had you just been integrating dmu from 0 to 1, because the latter has to agree with the Gauss' law result for the field from a point charge Q seen at the radius of the sphere. You see that the actual integral of mu*dmu is just 1/2 of that. So the field at the center of a hemisphere of charge is 1/2 the field at that same distance from a point charge, and the latter you can use Gauss' law for. Of course, if you know the field of a point charge without referring to Gauss' law, that saves a step.