# Electric field at centre of a hollow hemisphere shell.

• uOEE
In summary, the conversation discusses a problem with solving a question using calculus. The person asked for help and was advised to consider the direction of a unit vector when integrating over a surface. They were also given advice to use symmetry to determine the direction of the net electric field and work with the component in that direction.
uOEE

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## The Attempt at a Solution

Please excuse the poor writing. I believe it should be legible enough, but if you have any questions, i'll clarify or rewrite it.

Please excuse the poor writing. I believe it should be legible enough, but if you have any questions, i'll clarify or rewrite it.

My steps for solving this was filling in the values into the formula given. Using a square differential element on the surface of the sphere, as well as spherical co-ordinates.

i'm sure i just have a small error somewhere in the math, but I'm not too sure where... I've redone the question 3 times.

These two sources, among others, show that my answer is wrong. Though I did my question differently, so i can't be sure where i went wrong.

http://www.personal.utulsa.edu/~alexei-grigoriev/index_files/Homework2_solutions.pdf

Welcome to PF!

Looks like you treated the unit vector ##\hat{r}## as a constant vector when you pulled it out of the double integral. Does the direction of ##\hat{r}## vary as you integrate over the surface? If so, you cannot treat it as a constant vector.

How would I treat the vector? My apologies. Calculus is not my strong suit.

How would I go about doing that? Calculus is not my stop suit. And also, thanks for the help.
TSny said:
Welcome to PF!

Looks like you treated the unit vector ##\hat{r}## as a constant vector when you pulled it out of the double integral. Does the direction of ##\hat{r}## vary as you integrate over the surface? If so, you cannot treat it as a constant vector.

Use symmetry to see which direction the net E field will point. Then just work with the component of E that is in that direction. (Project your integrand into that direction.)

## 1. What is the formula for calculating the electric field at the centre of a hollow hemisphere shell?

The formula for calculating the electric field at the centre of a hollow hemisphere shell is E = kQ/R2, where E is the electric field, k is the Coulomb's constant, Q is the total charge on the hemisphere shell, and R is the radius of the hemisphere shell.

## 2. How does the electric field at the centre of a hollow hemisphere shell compare to the electric field at the centre of a solid hemisphere?

The electric field at the centre of a hollow hemisphere shell is zero, while the electric field at the centre of a solid hemisphere is non-zero. This is because the charge is distributed on the surface of a hollow hemisphere shell, while it is distributed throughout the volume of a solid hemisphere.

## 3. Can the electric field at the centre of a hollow hemisphere shell be negative?

No, the electric field at the centre of a hollow hemisphere shell cannot be negative. This is because the electric field is always directed away from positive charges, and since there are no charges at the centre of the hemisphere shell, the electric field is zero.

## 4. Does the electric field at the centre of a hollow hemisphere shell depend on the thickness of the shell?

No, the electric field at the centre of a hollow hemisphere shell does not depend on the thickness of the shell. As long as the total charge and radius of the shell remain the same, the electric field at the centre will also remain the same.

## 5. How can the electric field at the centre of a hollow hemisphere shell be used in real life applications?

The electric field at the centre of a hollow hemisphere shell can be used to study the behavior of electric charges and to design and improve electronic devices. It can also be used in electrostatic shielding to protect sensitive equipment from external electric fields.

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