Electric field of half a spherical shell

[asi123]

Homework Statement

Hey guys.
I've been trying to solve this question using Gauss law but I can't think of a surface that can contain this thing.
Is there another way to solve this?

Thanks a lot.

Homework Equations

The Attempt at a Solution

 

[CompuChip]

Maybe it helps to draw a picture... suppose that the symmetry axis is the z-axis. Consider an intersection in the (x, y) plane, i.e. a circle of radius r(z) at height z. Draw some electric field lines to a point on the z-axis... the symmetry will provide some cancellations. If you use an angle [itex]\phi[/itex] to parametrize the circle, then you can divide the circle into small segments of length [itex]r(z) \, d\phi[/itex] and write down the contribution to the E-field from each segment. Once you have that, all you have to do is sum (i.e. integrate) over phi.

Yes I know, it's a lot of work :sad:

 

[thomate1]

Dear student,

Thank you for reaching out for help with your question. The electric field of a half spherical shell can indeed be solved using Gauss' law. The key is to choose a Gaussian surface that encloses the charge distribution of the half spherical shell. In this case, a good choice would be a hemisphere with its flat side facing the half spherical shell. This surface would enclose all of the charge within the half spherical shell and make the calculations much simpler.

Another way to solve this problem is to use the concept of symmetry. Since the charge distribution is symmetric about the center of the half spherical shell, the electric field at any point on the surface of the half spherical shell will have the same magnitude and direction. This means that we can use the electric field equation for a point charge, E = kq/r^2, where k is the Coulomb's constant, q is the charge of the half spherical shell, and r is the distance from the center of the half spherical shell to the point of interest on the surface.

I hope this helps in your understanding of the electric field of a half spherical shell. Don't hesitate to reach out if you have any further questions. Good luck with your studies!

Best,

Scientist