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Electric Field due a charged disk

  • Thread starter Yalanhar
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  • #1
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Homework Statement:

uniformly charged disk, radius r, with surface charge density ##\sigma##
. I want to find the electric field along the axis through the centre of the disk at a h distance

Relevant Equations:

##dE=\frac {kdq}{r^2}##
Homework Statement: uniformly charged disk, radius r, with surface charge density ##\sigma##
. I want to find the electric field along the axis through the centre of the disk at a h distance
Homework Equations: ##dE=\frac {kdq}{r^2}##

My Solution:
##dE=\frac {kdq}{r^2}##
in this case r=s
##dE=\frac {kdq}{s^2}##

##dq=\sigma dA## where: ##dA=2\pi rdr##
##dq={\sigma 2\pi rdr }##

forum.png


##dE=\frac {1}{4\pi \epsilon_{o}}\frac {2\sigma \pi rdr}{s^2}cos\theta##
##dE=\frac {1}{2 \epsilon_{o}}\frac {\sigma rdr}{s^2}cos\theta## (1)

in the triangle:
##tan\theta = \frac {r}{h}## therefore : ##r = htan\theta## (2) and ##dr=\frac{hd\theta}{cos^2\theta}##(3)
##cos\theta = \frac {h}{s}## therefore : ##s = \frac {h}{cos\theta}##(4)

(2),(3) and (4) in (1)

##dE=\frac {1}{2 \epsilon_{o}}\frac {\sigma htan\theta (\frac{hd\theta }{cos^2\theta}) cos\theta }{(\frac {h}{cos\theta})^2}##
##dE=\frac{\sigma sin\theta d\theta}{2\epsilon_{0}}##
##E=\frac {\sigma}{2\epsilon_{o}}\int_0^\theta sin\theta \,d\theta##

So:
##E=\frac{\sigma}{2\epsilon_{o}}(1-cos\theta)##

Using ##cos\theta=\frac{h}{\sqrt {r^2 + h^2}}##

##E=\frac{\sigma}{2\epsilon_{o}}(1-\frac{h}{\sqrt{ r^2 + h^2}})##

Is it correct? Normally students don't do this way, so I am not sure. Also, why in the integration I don't need to multiply by 2? ##\theta## isn't only for the first half?
 
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Answers and Replies

  • #2
BvU
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Looks ok. Check e.g. here
(or anywhere googling sends you .... :rolleyes: )
 
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  • #3
haruspex
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I don't need to multiply by 2? θθ\theta isn't only for the first half?
You considered a complete annulus radius r (area 2πrdr). No part of the disc has been left out in taking θ from 0 to its max.
 
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