Voltage at a Point: Calculate Electric Field

[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.

 

[Redbelly98]

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

 

[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?

 

[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?

 

[forty]

Yerp, confusing sh*t like usual >.<

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

 

[Redbelly98]

Why are you talking about components?...The potential is a scalar not a vector ...

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