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 Mar 20th 2009, 01:18 PM #1 Senior Member   Join Date: Jul 2008 Posts: 179 Electric Fields A disk of radius equal to 10cm is loaded with a uniform density of charge $\displaystyle \alpha = 3x10^{-10}C/m^2$. Calculate the electric field at a distance of 1cm from the disk, in a point on its axis.
 Mar 20th 2009, 10:10 PM #2 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 Are you aware of the formula for the electric field due to a charged ring at its axis given by- E = xQ/4IIe(R^2 + x^2)^3/2 { II stands for Pi and e for epsilon-not) If yes, we can start from here and can find the field due to disk by integration the way we find moment of inertia of the disk. Please let me know.
 Mar 21st 2009, 07:57 AM #3 Senior Member   Join Date: Jul 2008 Posts: 179 Sorry, this formula is that most do not know
 Mar 21st 2009, 08:23 AM #4 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 Ok, which grade question is this. I mean which class you are studying in ? This might be a question of field due to large charged sheet given be Electric field = Surface charge density/ twice of epsilon not. But this will be a crude approximation.
Mar 23rd 2009, 09:25 AM   #5
Senior Member

Join Date: Jul 2008
Posts: 179
 Originally Posted by Parvez Ok, which grade question is this. I mean which class you are studying in ? This might be a question of field due to large charged sheet given be Electric field = Surface charge density/ twice of epsilon not. But this will be a crude approximation.

I am studying physics 2
- I not have all the data for this formula ?

 Mar 23rd 2009, 10:20 AM #6 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 The exact formula for the electric field at the axis of a charged disk is given by- E = k*2II*R^2*a/{ x*sqrt(R^2 + x^2)}. Here k =9*10^-9 Nm^2/C^2, II stands for Pi (3.14), R = 10 cm = 0.1 m, a is used for alpha and = 3*10^-10 C/m^2 and x = 1 cm = 0.01 m.
 Mar 23rd 2009, 10:24 AM #7 Senior Member   Join Date: Jul 2008 Posts: 179 Thanks for the explanations and the formula
 May 15th 2009, 07:01 PM #8 Senior Member   Join Date: Jul 2008 Posts: 179 I use the formula: $\displaystyle E = \frac{ \alpha}{2e.}(1-\frac{z}{\sqrt{z^2+R^2}})$ I not find the answer that is $\displaystyle 1,53x10^5N/C$ Last edited by Apprentice123; May 16th 2009 at 08:43 AM.
 May 17th 2009, 01:20 PM #9 Senior Member   Join Date: Jul 2008 Posts: 179 I find: 15,27N/C. But the answer is 1,53x10^5 N/C

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