1. A large, flat, nonconducting surface carries a uniform surface charge density

Question

1. A large, flat, nonconducting surface carries a uniform surface charge density . A small circular hole of radius R has been cut in the middle of the sheet.

Ignore fringing of the field lines around all edges and calculate the potential at point P, a distance z from the center of the hole along its axis. Express your answer in terms of , R, and z. (Hint: This is an example of when you can't define V = 0 to be at infinity; this issue arises whenever your charge distribution itself extends to infinity. So, in this problem, define the potential to be zero at the plane.)

(Note: Evaluate the integral, and show your work--just showing the result of the integral is not enough. I will accept results from tables of integrals, or from computer programs, though you must show the work that precedes that step, and clearly indicate what resource you used, and which part of your calculations come from that resource. Simplify your results as far as possible)

Explanation / Answer

Suppose that there is no hole at the center.

So the field at the required point P is /2

Potential at the required point =(-/2)*z

Now we have to calculate the potential of the disk which we have to remove at the center

Charge on small area of Small disk of radius x from the center of the sheet=dq = 2xdx

Potential due to this small part on the point P=Kdq / (z^2+x^2)=K*2xdx /  (z^2+x^2)

Now integrate it from 0 to R

This is very simple integration

Check it out in the link below

http://integrals.wolfram.com/index.jsp?expr=x%2Fsqrt%28z^2%2Bx^2%29&random=false

So After puting the integrating limits ,total potential dute to the disk which we suppose to be present at the center

=K*2*((R^2+z^2)-(z^2))

So the potential at point P=(-/2)*z - K*2*((R^2+z^2)-(z^2))

I worked hard on this problem.If the answer is not correct.Please give me some points for this answer please.

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