electric field problems


by eku_girl83
Tags: electric, field
eku_girl83
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#1
Jan24-04, 02:44 PM
P: 90
Here an electric field problems I'm struggling with:
Positive charge Q is uniformly distributed around a semicircle of radius a. Find the electric field at the center of curvature P. Answer in terms of Q, k, and a.
When I worked this out on my own, I calculated the magnitude of the electric field to be (2Qk)/(a^2), but this is wrong. Could someone give me a hint on how to do this correctly?
Thanks!
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Doc Al
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#2
Jan24-04, 03:11 PM
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Find the field for a segment (dΘ) of the semi-circle and integrate. Luckily it's an easy integral to do (since the distance is the same for all points on the curve). Try and set it up.
himanshu121
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#3
Jan24-04, 03:11 PM
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Use symmetry
Consider an axis passing through the Centre and dividing the semicircle into two halves So charge density(lambda)=Q/(pi)r.

Now consider dq charge symmetrical to axis u will see one component of Field produced cancels and other adds up in one direction. Can u show what u get from above hint

Anyway it is going to be moved to HW

eku_girl83
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#4
Jan24-04, 03:46 PM
P: 90

electric field problems


Does mean dE=(kdQ)/(a^2)?
dQ=Pi(a)(lamda) da?
Is this remotely close?
Doc Al
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#5
Jan24-04, 04:02 PM
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Originally posted by eku_girl83
Does mean dE=(kdQ)/(a^2)?
dQ=Pi(a)(lamda) da?
Is this remotely close?
[tex]dQ = \frac {Q}{\pi a} a d\Theta = \frac{Q d\Theta}{\pi}[/tex]
[tex]dE = \frac {k}{a^2} \sin \Theta dQ = \frac{Qk}{\pi a^2} \sin \Theta d\Theta [/tex]


I did neglect to mention, as himanshu points out, that you will need to take advantage of symmetry. Imagine the semicircle intersecting points (-a,0) (0,a) and (a,0). The only component of field you need to worry about is the y-component: by symmetry, the x-components (from opposite sides) cancel.
eku_girl83
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#6
Jan24-04, 04:13 PM
P: 90
I get (Qk)/(Pi a^2) directed down?
Thank you so much for helping me!
Doc Al
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#7
Jan24-04, 08:09 PM
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Originally posted by eku_girl83
I get (Qk)/(Pi a^2) directed down?
Thank you so much for helping me!
Check your work. I think you are off by a factor of 2. (Unless I made a mistake.) Did you integrate over the full range of Θ?
himanshu121
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#8
Jan25-04, 01:57 AM
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Yes u are off by a factor 2


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