# Potential and electric field.

#### Kruum

1. The problem statement, all variables and given/known data

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.

3. The attempt at a solution

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

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?

Last edited by a moderator:

#### tiny-tim

Homework Helper
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

Homework Helper
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

Homework Helper
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

Homework Helper
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

Homework Helper
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)

#### Kruum

no, no ln …

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

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

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

#### tiny-tim

Homework Helper
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 $$\frac{ \sigma}{2 \epsilon _0x}( \sqrt{x^2+R^2}-x)$$

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