Electric field due to a uniformly charged disc

  • Thread starter vinzie
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  • #1
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At the end of the derivation, it is given
The electric fiel due to elementary ring at the point P is dE = [2∏rσdrx]/[4∏epsilon zero (x^2 +r^2)^(3/2)
]

To find the total E due to disc is given by

∏σx/4∏ε(2rdr)/(x2 + r2)3/2

I am stuck with the calculus done here to arrive at the solution. Please help me.

Thank You!

Vinzie
 

Answers and Replies

  • #2
647
3
Do you mean

[tex]\mathrm{d}E=\dfrac{2\cdot\pi\cdot r\cdot\sigma\cdot x\cdot\mathrm{d}r}{4\cdot\pi\cdot\epsilon_0\cdot\left(x^2+r^2\right)^{\frac32}}[/tex]

and you want to know how to determine that

[tex]E=\dfrac{\pi\cdot\sigma\cdot x}{4\cdot\pi\cdot\epsilon_0}\cdot\displaystyle\int\left(\dfrac{2\cdot r\cdot\mathrm{d}r}{\left(x^2+r^2\right)^{\frac32}}\right)[/tex]

Also, why isn't the first TeX parsing?

Simple. Integrate both sides.
 
  • #3
5,085
123
I can't see your formula!!!!
 
  • #4
647
3
Yep, for some reason, the TeX isn't parsing. It should! It does on AoPS!
 
  • #5
19
0
Do you mean

[tex]\mathrm{d}E=\dfrac{2\cdot\pi\cdot r\cdot\sigma\cdot x\cdot\mathrm{d}r}{4\cdot\pi\cdot\epsilon_0\cdot\left(x^2+r^2\right)^{\frac32}}[/tex]

and you want to know how to determine that

[tex]E=\dfrac{\pi\cdot\sigma\cdot x}{4\cdot\pi\cdot\epsilon_0}\cdot\displaystyle\int\left(\dfrac{2\cdot r\cdot\mathrm{d}r}{\left(x^2+r^2\right)^{\frac32}}\right)[/tex]

Also, why isn't the first TeX parsing?

Simple. Integrate both sides.
Hi,

WOuld you mind if you tell me how to solve the integration part?

Thank You
 
  • #6
5,085
123
Divid [itex](x^2+r^2) \;[/itex] by [itex]r^2\;[/itex] to get [itex] 1+(\frac x r)^2\;[/itex] and use trig function to substitute. and see what happen. Something like [itex]\tan \theta=\frac x r [/itex]. [itex]1+\tan^2\theta=\sec^2\theta[/itex]
 
Last edited:
  • #7
19
0
Divid [itex](x^2+r^2) \;[/itex] by [itex]r^2\;[/itex] to get [itex] 1+(\frac x r)^2\;[/itex] and use trig function to substitute. and see what happen. Something like [itex]\tan \theta=\frac x r [/itex]. [itex]1+\tan^2\theta=\sec^2\theta[/itex]
Thanks Yungman!!

I am gonna solve that way.
 

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