A flat, insulating disk of radius R lies in the xy plane and is centered
at the origin. It has uniform surface charge density σ.

(a) Find the electric potential on the axis of the disk.

(b) Find the limiting behavior of the potential when z >> R,
and z << R.

Solution:

- Concepts:

The electric potential - Reasoning:

Φ(**r**) = k ∫_{A' }σ(**r'**)dA'/|**r**-**r**'|.

We are asked to find the potential due to a surface charge distribution in a situation with symmetry. - Details of the calculation:

dV(z) = k2πrσdr/(r^{2 }+ z^{2})^{½}is the potential on the axis due to a ring of radius r.

Here k = 1/(4πε_{0}).

Integrating from r = 0 to r = R we find V(z) = k2πσ{(R^{2}+z^{2})^{½ }- z} for a disk of radius R.

Q = σπR^{2}, so V(z) = k(2Q/R^{2}){(R^{2}+z^{2})^{½ }- z}.

If z >> R then V(z) » (2kQ/R^{2}){z(1 + R^{2}/2z)^{ }- z} = kQ/z. This is the potential of a point charge. From very far away the disk looks like a point.

If z << R then V(z) = (2kQ/R^{2}){R^{ }- z} = constant - 2kQz/R^{2}. The constant is arbitrary.

E_{z}= -dV(z)/dz = 2kQ/R^{2}= σ/(2ε_{0}). This is the field of an infinite plane with surface charge density σ.