Go Back   Physics Help Forum > College/University Physics Help > Advanced Electricity and Magnetism

Advanced Electricity and Magnetism Advanced Electricity and Magnetism Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Mar 20th 2009, 02:18 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.
Apprentice123 is offline   Reply With Quote
Old Mar 20th 2009, 11:10 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.
Parvez is offline   Reply With Quote
Old Mar 21st 2009, 08:57 AM   #3
Senior Member
 
Join Date: Jul 2008
Posts: 179
Sorry, this formula is that most do not know
Apprentice123 is offline   Reply With Quote
Old Mar 21st 2009, 09:23 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.
Parvez is offline   Reply With Quote
Old Mar 23rd 2009, 10:25 AM   #5
Senior Member
 
Join Date: Jul 2008
Posts: 179
Originally Posted by Parvez View Post
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 ?
Apprentice123 is offline   Reply With Quote
Old Mar 23rd 2009, 11:20 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.
Parvez is offline   Reply With Quote
Old Mar 23rd 2009, 11:24 AM   #7
Senior Member
 
Join Date: Jul 2008
Posts: 179
Thanks for the explanations and the formula
Apprentice123 is offline   Reply With Quote
Old May 15th 2009, 08:01 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 09:43 AM.
Apprentice123 is offline   Reply With Quote
Old May 17th 2009, 02:20 PM   #9
Senior Member
 
Join Date: Jul 2008
Posts: 179
I find: 15,27N/C. But the answer is 1,53x10^5 N/C
Apprentice123 is offline   Reply With Quote
Reply

  Physics Help Forum > College/University Physics Help > Advanced Electricity and Magnetism

Tags
electric, fields



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Electric Fields Question? emily36 Electricity and Magnetism 0 Jan 13th 2013 01:22 PM
Capacitors and electric fields ZergLurker Advanced Electricity and Magnetism 2 Feb 24th 2010 01:55 AM
Electric fields LS09 Electricity and Magnetism 1 Feb 6th 2009 10:20 PM
electric fields physics123456 Advanced Electricity and Magnetism 1 Jan 18th 2009 07:37 PM
Sketching the electric fields soccerdude28 Advanced Electricity and Magnetism 0 Nov 1st 2008 12:58 PM


Facebook Twitter Google+ RSS Feed