How Do You Calculate Electric Potential and Field on a Circular Plate's Axis?

[Kruum]

Homework Statement

Charge is uniformly distributed along a circular plate (radius R). surface charge is \sigma.
a) Determine the electric potential on the plates axis distance x from the center of the plate.
b) Calculate the electric field on the axis distant >x from the center of the plate.

The Attempt at a Solution

http://www.aijaa.com/img/b/00132/3756420.jpg
http://www.aijaa.com/img/b/00540/3756421.jpg

I couldn't get the E field any "cleaner" and in a I wasn't sore about the substitution of A (i.e. can I do it like that). fire my methods correct?

And don't mind my handwriting, MS Paint isn't that responsive.

Edit: Holy crap! Can I resize the images?

 

[tiny-tim]

I couldn't get the E field any "cleaner" and in a I wasn't sore about the substitution of A (i.e. can I do it like that). fire my methods correct?

Hi Kruum!

(have an integral: ∫ and a square-root: √ and a pi: π and an epsilon: ε and a sigma: σ )

In a), yes that's the way to get dA (but haven't you lost the 2 in 2πr dr?);

but your integral is wrong …

you should be able to integrate that just by looking at it, but if you can't, try susbtituting u = √(x² + r²)

 

[Kruum]

Thanks for having such fast fingers, tiny-tim!

In a), yes that's the way to get dA (but haven't you lost the 2 in 2πr dr?);

You can call me stupid, but isn't the area of a circle \pi r^2, hence my \pi rdr. So wouldn't 2 \pi rdr be the volume of a sylinder?

but your integral is wrong …

I can't believe it! I missed the minus... And the 2 should be 8.

 

[tiny-tim]

You can call me stupid, but isn't the area of a circle \pi r^2, hence my \pi rdr.

Yes it is …

but you want the area of a ring, of thickness dr ​

I can't believe it! I missed the minus...

uhh? what minus?

 

[Kruum]

Yes it is …

but you want the area of a ring, of thickness dr ​

Of course! Thanks...

uhh? what minus?

Nevermind... I've been recapping too much, I can't think clearly anymore! I rememberd that s=√x²-r².

But that electric field equation cannot be simplified?

Need... to... take... a... break...! :zzz:

 

[tiny-tim]

Nevermind... I've been recapping too much, I can't think clearly anymore! I rememberd that s=√x²-r².

Looks like s=√x²+r² to me …

anyway, try that substitution I mentioned ​

 

[Kruum]

anyway, try that substitution I mentioned

Isn't it \frac{ \sigma }{2 \epsilon _0} \int_0^R \frac{r}{ \sqrt{x^2+r^2}}dr= \frac{ \sigma }{4 \epsilon _0} ln( \frac{ \sqrt{x^2+r^2}}{x})

 

[tiny-tim]

Isn't it \frac{ \sigma }{2 \epsilon _0} \int_0^R \frac{r}{ \sqrt{x^2+r^2}}dr= \frac{ \sigma }{4 \epsilon _0} ln( \frac{ \sqrt{x^2+r^2}}{x})

Nope!

 

[Kruum]

Okay, now I'm completely lost! Can you point out where the error is? I've been using the fact that \int \frac{f'}{f}=ln(f)+C. There is an ln in the answer, right?

 

[Kruum]

I used the substitution method and got \frac{ \sigma }{2 \epsilon _0} \int_0^R \frac{r}{ \sqrt{x^2+r^2}}dr= \frac{ \sigma }{2 \epsilon _0} ln( \frac{ \sqrt{x^2+R^2}}{x}).

 

[tiny-tim]

Okay, now I'm completely lost! Can you point out where the error is? I've been using the fact that \int \frac{f'}{f}=ln(f)+C. There is an ln in the answer, right?

no, no ln …

hmm … ∫2r dr/(x² + r²) = ln(x² + r²) …

(and anyway ln(√(x² + r²)/x) = (1/2)ln(x² + r²) - lnx)

what's your f?

 

[Kruum]

no, no ln …

hmm … ∫2r dr/(x² + r²) = ln(x² + r²) …

(and anyway ln(√(x² + r²)/x) = (1/2)ln(x² + r²) - lnx)

what's your f?

Ignore that. So there's no ln in the answer? I'm just too tired to think anymore...

 

[tiny-tim]

get some sleep :zzz: …

 

[Kruum]

Looks like some food and a look at my maths book helped clear my mind... Now I get from the integral \frac{ \sigma}{2 \epsilon _0}( \sqrt{x^2+R^2}-x). So the E field would be \frac{ \sigma}{2 \epsilon _0x}( \sqrt{x^2+R^2}-x)