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.

the final answer is:

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

Quote:

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Originally Posted by braddy

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.

the final answer is:

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

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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

You assume right about the rho thing!!
Ok i will stick with my answer. I have some questions though. I will post them in a new post.
B