Calculating Electric Field at the Center of a Hemispherical Shell of Charge

In summary, the problem is finding the E field due to a hemispherical shell of charge. The solution involves using spherical coordinates and integrating from 0-->π.
  • #1
swervin09
7
0

Homework Statement


A hollow sphere of outer radius R2 and inner radius of R1 carries a uniform charge 2Q. The sphere is then cut in half to create a hemispherical shell of charge Q. Calculate E at the center point (origin) P.


Homework Equations


equation of a hollow sphere = 2/3π(r2-r1)
Gauss' Law ∫E dot dA
surface area hemisphere = 2πr^2


The Attempt at a Solution


Well, I know this is an integration problem and that I am better off integrating with polar coordinates and that I will be integrating from 0-->π as my lower and upper integral bounds.
But in all honesty I haven't had much fortune setting the integral up. The set up is the help I am asking for.
 
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  • #2
If I am not mistaken, you need to find the E field due to a HEMISPHERICAL shell

That involves some somewhat-complicated multiple integrals and E form Coulumb's law. Gauss's law will not work due to lack of symmetry
 
  • #3
Assuming that that is your problem, here is my hint:

Use spherical coordinates. find the charge density. consider an infinitesimal piece of the shell and the Coulomb force on a test charge at the origin. Then choose appropriate limits for r, theta, and phi and integrate
 
  • #4
Yes that is the problem. spherical coordinates make more sense. I will try that and post tomorrow what I have come up with. I don't have my calculus book nearby to refresh my memory of spherical coord. integration. This is one of those problems that has me intrigued and eager to "beat". It isn't worth a lot of points but it is due Thursday.
Thank you for the hint!
 
  • #5
no problem!

I might add: depending on how comfortable you are with multivariable, you don't have to use a triple integral; just find the field of a ring, sum into a washer, sum the washers into a shell. The spherical coordinates are just a way of thinking, no need to get formal about it
 
  • #6
I follow you up until you state sum the washers into a shell. Please provide a hint as to that specific.
 

Related to Calculating Electric Field at the Center of a Hemispherical Shell of Charge

1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. It is named after the mathematician and physicist Carl Friedrich Gauss.

2. How is Gauss' Law represented mathematically?

Gauss' Law is represented as an integral equation: ∮E⋅dA = Q/ε₀, where E is the electric field, dA is a small area element on the closed surface, Q is the enclosed charge, and ε₀ is the permittivity of free space.

3. What is the role of integration in Gauss' Law?

Integration is used in Gauss' Law to calculate the total electric flux through a closed surface by summing up the contributions from each small area element on the surface. This allows us to determine the total charge enclosed within the surface.

4. How is Gauss' Law applied in practical situations?

Gauss' Law is used in various practical applications, such as calculating the electric field of a charged particle, determining the electric field inside a charged conductor, and analyzing the behavior of electric fields in different situations.

5. What are the limitations of Gauss' Law?

Gauss' Law is based on the assumption of a static electric field and does not take into account the effects of changing magnetic fields. It also does not apply to systems with non-uniform charge distributions or those with moving charges. Additionally, it is only valid in scenarios with perfect conductors and does not consider the effects of resistance.

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