# Calculate the flux through the surface?

## Homework Statement

A hemispherical surface with radius e in a region of uniform electric field E has its axis aligned parallel to the direction of the field. Calculate the flux through the surface?

Flux = E A cos β

## The Attempt at a Solution

Why can't we say that the flux is E * Area of hemisphere which is E (2∏r^2)+(∏r^2) ???

Thanks for clarifying!

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NascentOxygen
Staff Emeritus
Why can't we say that the flux is E * Area of hemisphere which is E (2∏r^2) + (∏r^2) ???
Do the quantities either side of your plus sign have identical units??

That aside, if you want to involve a constant flux and the area of a hemisphere then you'd need that the flux be uniform over that area. But is isn't. Some of the hemisphere's surface is almost parallel to the E field, while some is almost perpendicular to E.

BTW, I think the hemisphere in question is 'open to the atmosphere', a surface comprising half a spherical shell.

Allright, I knew that the answer should be E (pi r^2) only... They said it should involve the down circle only... But my question is why? ... And a reply to your is that yes, the electric field should be constant over a hemisphere since the radius is the same everywhere... Hence, the flux should so be the same everywhere!!

tiny-tim
Homework Helper
hi ehabmozart!

hint: suppose it was a closed surface (the hemisphere plus the base) …

what would the flux be then?

Hi tinytim! :)

I guess your hint was the question itself... I consider the answer to be E * (2 pi r^2 + pi r^2) and not as the book suggests E pi r^2 :) Kindly help!

tiny-tim
Homework Helper
hmm …

how generally would you calculate the flux through a closed surface?

any theorems?

All what i've take is that Flux = E A ... We didn't even reach gauss's law so i want an easy approach to this!

can i get more clarification on why the electric flux through the circle will be the same which passes through the top part of the hemisphere?

haruspex
Homework Helper
Gold Member
can i get more clarification on why the electric flux through the circle will be the same which passes through the top part of the hemisphere?
Since the field is uniform and normal to the circle, every bundle of flux that passes through the hemispherical shell will pass through the circle, and vice versa.

Ok... that's fine.. But y?? I mean the circle is not having the same surface area as the hemisphere does?

NascentOxygen
Staff Emeritus
Sketch a field of parallel flux lines, and draw in a hemisphere. (No need to post it here, so don't worry about not having good sketching ability.)

Count the number of flux lines that pass through the hemispheric surface. Compare this count with the number of flux line that pass through the flat circular surface. Can you see any relationship?

tiny-tim
Homework Helper
(just got up :zzz:)
can i get more clarification on why the electric flux through the circle will be the same which passes through the top part of the hemisphere?
hint: what is the divergence (at a general point (x,y,z)) of an electric field of constant magnitude E in the z direction?

The reason that it is E*(pi*r^2) isn't because only the bottom part matters. It's because, in the top part, the x-and y-components of flux cancel out, due to the n^ vectors coming out of the surface at varying directions, leaving only the contributions of the positive z-direction n^ vectors, and in the bottom, all n^ vectors point in the -z-direction. Thus, the net of the flux would be E(surface area of top part)-E(surface area of base)=E(2*pi*r^2)-(pi*r^2), leaving just E(pi*r^2).

Fab
Since the E field goes inward on the circular base, the flux is -E(pi*r^2). Since it is pointing outward from the concaved part, the flux is E(2pi*r^2) (since it is half a sphere the area is halfed). Then just compine the two