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a charged segment of a ring?

 
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Feb25-13, 11:58 PM   #1
 

a charged segment of a ring?


Hi everyone, i have been looking at this problem for a few days now and i am pretty stuck, ill explain why

first the problem :
a charge Q is distributed over a ring section of radius r, between angles 90 and 180. find the V and E at the origin.

Now the problem im having ais everything that is kind of similar to this problem or about charged rings is only for uniformly charged rings and i cant find anything about just a segment, and in those cases i think E at the center is 0. Also every where i look im seeing a different formula for E of a charged ring. I am pretty confused
 
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Feb26-13, 08:27 AM   #2
 
i also dont understand how only one segment and in this case the 2nd quadrant can be charged..
 
Feb26-13, 12:18 PM   #3
 
[tex]\vec{E}=\int\frac{\lambda\hat{r}}{r^2}ds[/tex]
Where [itex]\lambda[/itex] merely represents the charge distribution [itex]Q[/itex], and [itex]\hat{r}[/itex] represents a unit vector perpendicular to each segment pointing towards the origin. You want to take the line integral on the interval [itex](\pi/2, \pi)[/itex].

Normally, the electric field would be zero in the center of a circle with uniform charge distribution. However, it is not in this instance.
 
Feb26-13, 11:30 PM   #4
 

a charged segment of a ring?


thank you, what source did you get that equation from id like to read more about it
 
Feb26-13, 11:57 PM   #5
 
i see 1/4*pi*eo and to the power of 3/2 in alot of ring equations thats was confusing me is the difference between those and this one
 
Feb27-13, 06:58 AM   #6
 
Not entirely sure what level you are at, but this was in both my vector calculus and electrostatics textbook. You may find variations depending on how the charge is distributed (line, surface, etc).
 
Feb27-13, 11:55 AM   #7
 
College, and the one in my Electromagnetics book is

E=[itex]\hat{z}[/itex][h/(4[itex]\pi[/itex]ε0(b^2+h^2)^3/2)]*Q where b is the radius and h is the height, but like i said its for a uniform charge
 
Feb27-13, 02:42 PM   #8
 
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Quote by Octavius1287 View Post
College, and the one in my Electromagnetics book is

E=[itex]\hat{z}[/itex][h/(4[itex]\pi[/itex]ε0(b^2+h^2)^3/2)]*Q where b is the radius and h is the height, but like i said its for a uniform charge
This is probably what it's referring to.



I don't know if this will be helpful, but take a look.

http://hyperphysics.phy-astr.gsu.edu...ic/elelin.html

Regards
 
Feb27-13, 07:13 PM   #9
 
ya all those make sense to me, but i read that page and once again it said it was only for a uniform charge
 
Feb27-13, 08:15 PM   #10
 
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Quote by Octavius1287 View Post
... i read that page and once again it said it was only for a uniform charge
Yes.

Quote by Octavius1287 View Post
...problem :
a charge Q is distributed over a ring section of radius r, between angles 90 and 180.
This doesn't mean it is NON-uniform for the rest of the ring. Aren't they looking for the field from the contribution of charge on just this piece of the ring?

If so,

Quote by sandy.bridge View Post
...You want to take the line integral on the interval [itex](\pi/2, \pi)[/itex].
 
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