Electric field due to a uniformly charged disc

[vinzie]

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

 

[Whovian]

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.

 

[yungman]

I can't see your formula!

 

[Whovian]

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

 

[vinzie]

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

 

[yungman]

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

 

[vinzie]

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 going to solve that way.