# Electric flux through a circular plate

1. Jul 30, 2008

### rolls

http://img508.imageshack.us/img508/7047/problemhi1.jpg [Broken]

Now I have gone about solving this problem by drawing a sphere with Q centred in it, lets call this sphere A1, now this circle defined by r=5 (lets call this circle C) cuts this sphere perfectly creating a spherical surface lets call As.

Now we know the electric flux through As will be the same as the area defined by the circle C so we can find the electric flux by finding the total electric flux through A and * it by the ratio of As/A.

So now we have: (sorry I don't know latex)
Ea1 = flux through A1 (sphere)
Eas = flux through As (spherical surface)

Eas = Ea1 * As/A1

A1 = 4*Pi*r^2
As = 2*Pi*r(r-d)
r = Sqrt(a^2 + d^2)
r = 5.83
Ea1 = Q/4*Pi*E0*r^2

Eas = Q/4*Pi*E0*r^2 * 2*Pi*r(r-d) / 4*Pi*r^2

Now I haven't bothered to simply it and calculate an answer but does this look correct?

Last edited by a moderator: May 3, 2017
2. Jul 30, 2008

### Defennder

Yeah it looks ok. Funny that they would actually give you the formula for surface area of a spherical cap and even tell you explicitly how to substitute the variables.

3. Jul 30, 2008

### rolls

Well it was the first question in the course, I've only have 1 lecture on the material so I really did need that hint.

4. May 24, 2011

### sanjanaraj

can anybody explain why flux is taken as (q/4phiepselon r^2) [which is the formula for field]and not as q/epsilon????please reply soon

5. May 24, 2011

### SammyS

Staff Emeritus
Probably not!!

As implied by your question, it looks like the given solution was in error - in just the way you suspect.

6. May 27, 2011

### sanjanaraj

consider a imaginary sphere S centered at Q,the circle cuts the sphere creating a spherical surface A.The flux through the circle is same as the flux through the spherical surface.
let the radii of the sphere subtend an angle 2(alpha) at the centre such that
tan(alpha) = a/d
now area of spherical surface,As=2*phi*a(1-cos(alpha) )
flux through the spherical surface = (q/epsilon)*2*phi*a(1-cos(alpha)) / {4*phi*(a^2 + d^2)
thus flux through the circular surface =
{q*a[1-(d/root(a^2+d^2)]}/{2*epsilon*(a^2+d^2)}

is it correct????

7. May 29, 2011

### SammyS

Staff Emeritus
I think AS should be given by:$$A_S=2\pi Rh=2\pi R(R-d)=2\pi R^2(1-d/R)=2\pi R^2(1-\cos(\alpha))=2\pi (a^2+d^2)(1-\cos(\alpha))\,.$$