# Electric field due to a uniformly charged disc

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

Thank You!

Vinzie

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Do you mean

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

and you want to know how to determine that

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

Also, why isn't the first TeX parsing?

Simple. Integrate both sides.

Yep, for some reason, the TeX isn't parsing. It should! It does on AoPS!

Do you mean

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

and you want to know how to determine that

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

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

Divid $(x^2+r^2) \;$ by $r^2\;$ to get $1+(\frac x r)^2\;$ and use trig function to substitute. and see what happen. Something like $\tan \theta=\frac x r$. $1+\tan^2\theta=\sec^2\theta$

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Divid $(x^2+r^2) \;$ by $r^2\;$ to get $1+(\frac x r)^2\;$ and use trig function to substitute. and see what happen. Something like $\tan \theta=\frac x r$. $1+\tan^2\theta=\sec^2\theta$
Thanks Yungman!!

I am gonna solve that way.