Electric field at the center of curvature of a solid hemisphere

[Aero6]

Homework Statement

Hi,
I'm wondering if this is the proper way to approach this problem. The question says to:

a)find the electric field at the center of curvature of the hemisphere (center of the flat bottom).

Homework Equations

Gauss's law: integral E*da = Qencl/epsilon

The Attempt at a Solution

used gauss's law and constructed a Gaussian surface around a solid sphere, found Qencl=k*pi* r^4
and then set integral E*da = Qencl/epsilon. I found the total electric field to be kr^2/8epsilon.

Since this was the electric field at the center of an entire solid sphere, I divided this answer in half since we only have the contribution of half of a solid sphere. I tried to check my answer by using gauss's law and integrating the polar angle from 0 to pi/4 for the hemisphere instead of 0 to pi for the whole sphere but got a different answer for the electric field. Did I get different answers because I cannot check the answer using gauss's law if I'm only integrating over half of a sphere?

Thank you.

 

[Delphi51]

I don't know how to do this one. Your answer isn't very good because the hemisphere is quite a different beastie from a sphere. The Gaussian surface around the sphere would give you the E field on the outside, curved part of the sphere whereas you are looking for E on the flat part of the hemisphere. Applying Gauss' law to the hemisphere is problematical because the E field would vary on this surface in a complex way. For example, it would likely be very strong near the circle where the curved surface meets the flat surface.

Do you have any information on how the charge is distributed in or on the hemisphere? That should be complex, too - something that the question writer might have simplified for you to make the problem solvable. For example, if the charge is distributed uniformly throughout the hemisphere you could fairly easily integrate over all the dq's to get the E at the spot you are interested in.

 

[Aero6]

The problem word for word is:
"A solid hemisphere of radius R has its center of curvature at the origin, with the positive z axis pointing out the 'dome' of the hemisphere. it has a charge density rho=alpha*r, with r measured from the origin.

a) find the electric field (magnitude and direction) at the center of curvature of the hemisphere (center of the flat bottom).

b) Find the electrostatic potential V a the same point, taking the potential to be 0 at infinity.

For part a) I want to try and use the formula for the electric field over a volume.
1/4*pi*epsilon*integral (1/r^2)*(r hat)*rho(r vector)*dtao but I am confused on how to figure out what r hat and the magnitude of r hat r. In our textbook, r vector is the separation vector, the vector between the charge and the point of interest. But often times we have to break up r vector into components and that's where I start to get lost. So I tried to use Gauss's law (in all honesty) to avoid using this formula.

 

[Delphi51]

Ah, much clearer now! Yes, you must integrate over the charge.
I don't recognize your fancy formula, but it must basically be good old
E = kQ/r^2. You probably use that pi and epsilon bit instead of my k.

The trick is to choose a small bit of the volume dV which has a charge dq = rho*dV for your integral. The brute force way would be to go with rectangular coordinates x,y and z and you'll have a 3D integral. There should be a shortcut, but I don't see it. You only need to consider the vertical (z) component of the E because the other components cancel out due to the symmetry.

 

[Delphi51]

I think the integral would be much easier if you considered a series of spherical surfaces with height dR to make a series of eggshells. That whole eggshell volume would be at distance R from the center point, which is a nice simplification. Of course the E vectors due to different parts of it would point in different directions, so that volume must be broken down further into a series of rings at angle of elevation A. The E due to that still has a range of directions but their z components will all be the same. This way you only have a 2D integral over A and then R.