- #1

Yalanhar

- 39

- 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

- 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 }##

##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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