# Homework Help: Voltage at a point.

1. Nov 12, 2008

### forty

http://students.informatics.unimelb.edu.au/serve/cmcleod/stuff/elec.JPG

Eo = permittivity of free space

integral dV = (1/4(pi)Eo) integral (dQ)/a

Hopefully I have that formula correct.

so....

dQ = 2(pi)r(dr)(sigma)

Horizontal components cancel. So there will be a sin(theeta) term at the end of the integral.

sin(theeta) = z/a

and a = sqrt(z^2 + r^2)

and the integral will be between 0 and R

so the RHS becomes...

(1/4(pi)Eo) integral(0 -> R) (2(pi)r(dr)(sigma)/sqrt(z^2 + r^2)) * z/sqrt(z^2 + r^2)

Is this right? I have a feeling I have done something stupid.

Any help greatly appreciated.

Last edited by a moderator: Apr 23, 2017
2. Nov 12, 2008

### Redbelly98

Staff Emeritus
I think that's correct. Since they give you the final answer, you can see if you get it when you do the integral.

3. Nov 12, 2008

### forty

once you pull all the constants out the front you get....

(z(sigma)/2Eo) integral(0 -> R) r/(z^2 + r^2) (dr)

I'm pretty sure that doesn't give that answer.

Any suggestions?

4. Nov 12, 2008

### gabbagabbahey

Why are you talking about components?...The potential is a scalar not a vector(perhaps you are confusing it with the electric field )....Do you really have a $\sin \theta$ term in your integrand?

Last edited: Nov 12, 2008
5. Nov 12, 2008

### forty

Yerp, confusing sh*t like usual >.<

But that all works out now! Thanks alot for pointing that out for me :)

6. Nov 13, 2008

### Redbelly98

Staff Emeritus
Oops, I missed that little detail Good save, ggh.