Charged Ring Over An Infinite Grounded Plane

We could use the method of image to solve it. Since we are dealing with infinite plane, then by partitioning the circular wire into very small elements, we conclude that the image is also a circular wire with radius $R$ and charge $-q$ located at distance $h$ below the plane.

The electric potential of a circular wire with charge $q$ and radius $R$ at the point located in a line passing through the center of the wire is given by $V(z) = \frac{q}{4\pi\epsilon_{0}} \frac{1}{\sqrt{R^{2} + z^{2}}}$ where $z$ is the distance between the point and the center of the wire.

Now, let take the origin at the center of the image circular wire and z-axis is the line that connecting the center of the wire and it's image. Using formula above, we could write the electric potential at the point located in z axis as, $V(z) = \frac{q}{4\pi\epsilon_{0}}\left(\frac{1}{\sqrt{R^{2} + (2h-z)^{2}}} - \frac{1}{\sqrt{R^{2} + z^{2}}}\right)$

And the induced surface charge, $\sigma = -\epsilon_{0} \left.\frac{\partial V}{\partial z}\right|_{z=h}$ or $\sigma = -\frac{qh}{ 2\pi\left(R^{2} + h^{2}\right)^{3/2}}$ and finally we have, $|\sigma |= 8.1\times 10^{-6}$ .

$LaTeX$ The method of images suits this case the best.Put an imaginary ring on the other side with a charge -q find the field due to this combination at the required point.By gauss theorem this must be equal to sigma/£°.