Answer:
(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:
[tex]V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }[/tex]
With [tex](\epsilon_{0})[/tex] been the vacuum permittivity
(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:
[tex]E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }[/tex]
Explanation:
(A) Considering a uniform linear density [tex]λ_{0} [/tex] on the ring, then:
[tex]dQ=\lambda dl[/tex] (1)⇒[tex]Q=\lambda_{0} 2\pi R[/tex](2)⇒[tex]\lambda_{0}=\frac{Q}{2\pi R}[/tex](3)
Applying the technique of charge integration for finite charges:
[tex]V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ[/tex](4)
Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.
Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:
[tex]r'=\sqrt{R^{2} +Z^{2}}[/tex](5)
Using the expressions (1),(4) and (5) you obtain:
[tex]V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi[/tex]
Integrating results:
[tex]V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }[/tex] (S_a)
(B) For the expression of the magnitude of the field E(z), is important to remember:
[tex]|E| =-\nabla V[/tex] (6)
But in this case you only work in the z variable, soo the expression (6) can be rewritten as:
[tex]|E| =-\frac{dV(z)}{dz}[/tex] (7)
Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):
[tex]E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }[/tex] (S_b)
