# Electric flux of a disk

1. Mar 12, 2009

### Dell

what is the electric flux through a disk with a radius R when a charge of q is placed at a distance of 0.5R perpendicular to the centre of the disk ?

2. Mar 12, 2009

### Vuldoraq

$$\int \underline{E}\bullet d\underline{A}=\phi_{E}$$

where $$d\underline{A}=dA*\widehat{n}$$ and n is the unit vector normal to the surface.

To visualise this try drawing your point charge, then drawing the disc below it, and draw the field you know the charge will produce.

3. Mar 12, 2009

### Dell

thanks, but thats how i would do it if i had a disk and a straight forward field, but how do i deal with a field that comes from a point charge R/2 away from the disk, since at every part of the disk there is a different flux since there is a different angle between n and E, how do i do this integral?

4. Mar 12, 2009

### Vuldoraq

Firstly you need to find what the electric field will be around the charge. Then consider the disc. It is perpendicular to a line joining it to the charge. So what will be the unit normal to the disc?

This problem has a lot of symmetry, so I would suggest using a coordinate system that exploits this, (try spherical polars).

5. Mar 12, 2009

### Vuldoraq

You can also consider the disc to be the top of a hemisphere. Since there is no source inside the hemisphere, the net flux must be zero. This means the flux through the disc is equal to the flux through the 'open' hemisphere.

In other words you can bend your disc into a hemisphere, with the same radius as the disc. This will make the E-field constant for your surface, so it can come outside the integral and then you are left with a trivial integral.

6. Mar 12, 2009

### Dell

welll the field around the charge will be Kq/r2, right?
but i dont see how that will help me,
the normal to the disk is 90 degrees to the disk and 90 to the charge, obviously,
dont really understand the rest of what youre saying though, how can i bend the disk, wont that change the flux, since i change the angle between the n and E verctors??

7. Mar 12, 2009

### Vuldoraq

Sorry, don't mean to confuse you. Gauss's law states (somewhere) that the flux through any closed surface is zero if that surface contains no source or sink. Now if you take your disc, and add to the bottom of it a hemisphere, then you will have created a closed surface that contains no sources or sinks. This means that the net flux through the entire surface is zero. Now the flux is coming in through the disc and out through the curved hemispere. Hence the flux through the disk is equivalent to the flux through the open hemisphere.

Now, your electric field is radial and the normal to the surface is also now radial. In addition the electric field is now constant over your new surface and so it can come outside the integral and all your left is the integral of dA for a hemisphere, which is just half the surface area of a sphere.

8. Mar 12, 2009

### Dell

okay, i think i understand,,,
so you mean, i take an imaginary "container" which is a half sphere "bowl" on one side and the "lid" is my disk. now the flux in through the bowl is my flux out through the lid ?

but how is the field constant over my new surface? surely the field can only be uniform if my charge is at the core of the spere, in which case the charge is inside the sphere.

if the charge is inside the shpere all our presumptions are incorrect since the flux is no longer 0

9. Mar 12, 2009

### Dell

if i say that the distance from the disk to the charge is 0.5R and the radius of the disk is R, then i can work out that the disk recieves 126.87 degrees of the field's possible 360 degrees ( if i could draw it you would understand) so it recieves (126/360)= 35% of what a sphere surrounding the charge would have recieved,

now i know that if it is a closed surface the size doesnt matter so lets take a sphere with a radius of R

flux= E*A = (Kq/R^2)*4pi/R^2 = 4K(pi)q

so the flux through my disk will be 0.35*4K(pi)q = 3.96*10^10

10. Mar 12, 2009

### Dell

my attempt at a diagram

let the red dot be the point chrage, and the green lines be its field lines,
trigonometry tells me that the angle between the 1st and last field line that touches the disk is 126.87 degrees

Last edited by a moderator: May 4, 2017
11. Mar 12, 2009

### Dell

now looking at it, that might be what you meant??
but im not getting it right,
the answer in my book is (6.2432*10^10)q ,i get (3.96*10^10 )q

Last edited: Mar 12, 2009
12. Mar 12, 2009

### Dell

okay, now i really think im there,(almost) could you just tell me where im going wrong,
here are my workings

i took a spherical surface with a radius of $$\frac{\sqrt{5}}{2}$$R which is flat on one side(the disk -green in my diagram) and said, as you advised, that since there is no internal charge, all the ingoing flux, (through the disk) is equal to the outgoing flux( through the spherical part)

now the area of the shere is 4$$\Pi$$rh which comes to
$$\frac{5-\sqrt{5}}{2}$$$$\Pi$$R2

$$\Phi$$=E*A
and i get
$$\Phi$$=$$\frac{q}{\epsilon}$$*$$\frac{5-\sqrt{5}}{10}$$

the answer in my book is
$$\frac{q}{4\epsilon}$$*$$\frac{\sqrt{5}-1}{\sqrt{5}}$$

my answer is half of the books, where am i going wrong??

Last edited by a moderator: May 4, 2017
13. Mar 12, 2009

### Vuldoraq

Well done on coming through the calculation, I'm impressed. My explanation was a little lacking. As for why your a factor of two out, I don't know. I have checked your calculations and they appear to be fine. I will keep looking.

14. Mar 12, 2009

### Dell

thanks, appreciated, did you come to the same answer?