E field of a hemisperical shell

[nosmas]

My teacher explained a problem of a hemispherical shell in class but i don't understand what he is doing.

http://img116.imageshack.us/img116/7656/naamloos27mf.gif

 

[tiny-tim]

welcome to pf!

hi nosmas! welcome to pf!

(he only worked out the z component, because from symmetry the x and y components must be zero)

he sliced the shell into rings because for any one ring, the z component of the field must be the same …

so if the total charge of that ring is q, then it has the same effect (on the z component) as a charge q all at one point (instead of spread out around the ring)

then he multiplied charge x 1/distance² x cosθ
​

(he seems to have unnecessarily put in a lot of r's that then canceled …

i expect that's because they were in Eq 23-10)

 

[nosmas]

Equation 23-10 is dE = k*(dQ/r^2)

What I am struggeling with is how the distance to the point of interest z = rcos(theta) and how they came up with the charge on the ring?

 

[tiny-tim]

hi nosmas!

What I am struggeling with is how the distance to the point of interest z = rcos(theta) and how they came up with the charge on the ring?

rcosθ is the distance from the centre to the plane of the ring

the charge is the charge density times the area,

and the area is the arc-length (rdθ) times the circumference of the ring (2πrsinθ)

 

[asteriatic]

The E field of a hemispherical shell can be calculated by using the formula for the electric field of a point charge. In this case, the hemispherical shell can be approximated as a collection of infinitesimal point charges, each with a small amount of charge. The total electric field at a point can then be found by adding up the contributions from each of these point charges using the principle of superposition.

In the image provided, your teacher is likely demonstrating how to use this principle to calculate the electric field at a point within the hemispherical shell. By dividing the shell into infinitesimal rings and calculating the electric field contribution from each ring, your teacher is essentially integrating over the entire surface of the shell to find the total electric field at the point of interest.

It is important to note that this calculation assumes the hemispherical shell is a perfect conductor, meaning the charge is distributed evenly on the surface and there are no internal electric fields. If the shell is not a perfect conductor, the calculation becomes more complex and may require additional considerations. I suggest discussing this further with your teacher or consulting a textbook for a more detailed explanation.