# Missing something

• Mar 3rd 2007, 12:02 PM
Missing something
Hi I think I am missing something in the solution of this problem.
I have the final solution of the problem but I get something a little bit different:

A circular disk of uniform surface integral "RHOs" and radius a lies in the x-y plane with center at (0,0,0). Find the potential PHI(z) along the z axis when the potential is Vo is at the center of the disk, using the potential formula.

PHI(z)=(RHOs/2*epsilon0)*[sqrt(a^2+z^2) -z-a]+Vo

I found:
PHI(z)=(RHOs/2*epsilon0)*[sqrt(a^2+z^2) -z]+Vo

I dont know how they get the variable a in bold.
Please can someone tell me what I get it wrong.
Thank you.
B
• Mar 3rd 2007, 03:39 PM
topsquark
Quote:

Hi I think I am missing something in the solution of this problem.
I have the final solution of the problem but I get something a little bit different:

A circular disk of uniform surface integral "RHOs" and radius a lies in the x-y plane with center at (0,0,0). Find the potential PHI(z) along the z axis when the potential is Vo is at the center of the disk, using the potential formula.

PHI(z)=(RHOs/2*epsilon0)*[sqrt(a^2+z^2) -z-a]+Vo

I found:
PHI(z)=(RHOs/2*epsilon0)*[sqrt(a^2+z^2) -z]+Vo

I dont know how they get the variable a in bold.
Please can someone tell me what I get it wrong.
Thank you.
B

What the heck does "uniform surface integral RHOs" mean? The Greek letter rho indicates a (volume) charge density, which I suppose I could assume the subscript "s" is meant to indicate. But the problem you are doing appears to be simply for a constant surface charge density (RHOs in your notation). What's this business about uniform surface integrals?

In any event, for a disk of uniform surface charge density, the electric potential for a point directly above the center of the disk is given by the formula that you derived, not what your given answer is. Now, by resetting the 0 point (ie. redefining the value of V0) we certainly CAN put it in the form that your given answer indicates (since your two solutions only differ by a constant), but I see no value in doing that.

That's a long winded way of saying I agree with your answer. :)

-Dan
• Mar 3rd 2007, 07:33 PM