Question

A disk of radius R has positive charge uniformly distributed over a inner circular region of radius A and negative charge uniformly distributed over the outer annular region. The surface density on the inner region is +o and that of the outer is - o. The electrostatic potential at P located on the central perpendicular axis is a distance z=R from the disk center is zero

What is A? In term of R

Answer 1

The potential from the outer ring:

$$V_p=\frac\sigma{2\epsilon_0}[R\sqrt{2}-\sqrt{R^2+(R-A)^2}].$$

The potential of the inner disk:

$$V_p=\frac\sigma{2\epsilon_0}(R\sqrt{2}-R).$$

For the field to be zero, these potentials must be equal:

$$\frac\sigma{2\epsilon_0}[R\sqrt{2}-\sqrt{R^2+(R-A)^2}]=\frac\sigma{2\epsilon_0}[R\sqrt{2}-R],\\\space\\ \sqrt{R^2+(R-A)^2}=R,\\ R^2+R^2-A^2-2AR=R^2,\\ A^2+2AR-R^2=0,\\\space\\ A=\frac{\sqrt{4R^2+4R^2}-2R}{2}=R(\sqrt2-1).$$