Projecting Area and Flux of Hemisphere, Frustum & Cone

[stronghold.mr]

PREVIOUS POST ----https://www.physicsforums.com/showthread.php?t=165844
THIS POST HELP ME OUT TO SOME EXTENT BUT I AM STILL MESSED UP...

I am messed up with the concept of projection of area of different surfaces and flux through different surfaces...

Please someone explain it with some examples such as -

1. Hemisphere (I AM GETTING --- FLUX = E pi r^2 ) PLEASE CONFIRM
2. Frustum (not getting)
3. Cone (not getting)

we have to find flux through all these surfaces (The field E is uniform)

 

[momo1111]

sorry about my English ...
- What came in must come out ...
3.
this is your Diagram:
(N.B. the dots in the cone are just to give it shape)

E
-->.../.\
-->../...\
-->./...\ <---- cone
-->/____\___________ <- table

...
E
-->.../!
-->../.!
-->./..! <---- half cone
-->/__! ___________ <- table
you have a triangle so the flux is :(surface of triangle) * (E) = ...
you can work out the same with othe problems .

 

[stronghold.mr]

thanks for ur reply

but can u please explain it for frustum and hemisphere

not getting the answer in case of frustum

and please confirm hemisphere one

thanks

 

[momo1111]

What is the orientation of the hemisphere ?

case 1 :
flux = (surface of semi Circle )*E
or
case 2 :
flux = (surface of a Circle )*E

 

[paranormal]

Sure, I would be happy to explain the concept of projecting area and flux through different surfaces.

First, let's define what we mean by "flux." Flux is a measure of the flow of a physical quantity through a surface. In this case, we are interested in the flow of an electric field (E) through different surfaces.

Now, let's look at the three surfaces you mentioned - hemisphere, frustum, and cone. In order to calculate the flux through these surfaces, we need to consider the projection of their area onto a plane perpendicular to the direction of the electric field.

1. Hemisphere:
To calculate the flux through a hemisphere, we need to consider the projection of its curved surface onto a plane perpendicular to the electric field. This projection will result in a circular area with a radius equal to the radius of the hemisphere (r). The flux through this circular area can be calculated using the formula you mentioned - Flux = E x projected area = E x πr^2. This is because the electric field is uniform and perpendicular to the plane, so the flux is simply the product of the field strength and the area.

2. Frustum:
A frustum is a cone with the top cut off. To calculate the flux through a frustum, we need to consider the projection of its curved surface onto a plane perpendicular to the electric field. This projection will result in a circular area with a varying radius (r) depending on the distance from the top of the frustum. The flux through this circular area can be calculated using the same formula as for the hemisphere - Flux = E x projected area = E x πr^2. However, in this case, r will vary depending on the distance from the top of the frustum. This means that the flux will also vary at different points on the surface.

3. Cone:
Similar to the frustum, to calculate the flux through a cone, we need to consider the projection of its curved surface onto a plane perpendicular to the electric field. This projection will result in a circular area with a varying radius (r) depending on the distance from the base of the cone. The flux through this circular area can also be calculated using the same formula as for the hemisphere - Flux = E x projected area = E x πr^2. However, in this case, r will vary depending on the distance from the base of the cone. Again, this means that the flux will also vary at different points on the