# Electric field of half a spherical shell

• asi123
In summary, the person is seeking help with a question that involves using Gauss law. They are struggling to find a surface that can contain the problem and are wondering if there is another approach to solve it. Another person suggests drawing a picture and considering the symmetry axis to aid in finding a solution. They also suggest dividing the circle into small segments and integrating over phi.
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.

## The Attempt at a Solution

#### Attachments

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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 $\phi$ to parametrize the circle, then you can divide the circle into small segments of length $r(z) \, d\phi$ 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:

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

## 1. What is the formula for the electric field of a half spherical shell?

The formula for the electric field of a half spherical shell is given by E = (Q/4πε0) * (sinθ2 - sinθ1), where Q is the charge of the shell, ε0 is the permittivity of free space, θ1 is the angle at the center of the shell, and θ2 is the angle on the surface of the shell.

## 2. How is the electric field of a half spherical shell different from that of a full spherical shell?

The electric field of a half spherical shell is only present on one side of the shell, while the electric field of a full spherical shell is present on both sides. Additionally, the electric field of a half spherical shell is only dependent on the angle at the center and on the surface of the shell, while the electric field of a full spherical shell is dependent on the distance from the center.

## 3. What is the direction of the electric field of a half spherical shell?

The direction of the electric field of a half spherical shell is always perpendicular to the surface of the shell. This means that the electric field lines will be directed away from the shell on one side and towards the shell on the other side.

## 4. How does the charge distribution of the half spherical shell affect the electric field?

The charge distribution of the half spherical shell affects the electric field by determining the strength of the electric field at different angles on the surface of the shell. If the charge is evenly distributed, the electric field will be constant at all angles. However, if the charge is unevenly distributed, the electric field will be stronger at certain angles and weaker at others.

## 5. Can the electric field of a half spherical shell be calculated using Gauss's law?

Yes, the electric field of a half spherical shell can be calculated using Gauss's law. This law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. By applying this law to a half spherical shell, the electric field can be calculated using the formula E = (Q/4πε0) * (sinθ2 - sinθ1).

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