Just as optical astronomers observe the visible light emitted by objects such as stars and galaxies, radio astronomers can also
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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:
With been the vacuum permittivity
(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:
Explanation:
(A) Considering a uniform linear density on the ring, then:
(1)⇒
(2)⇒
(3)
Applying the technique of charge integration for finite charges:
(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:
(5)
Using the expressions (1),(4) and (5) you obtain:
Integrating results:
(S_a)
(B) For the expression of the magnitude of the field E(z), is important to remember:
(6)
But in this case you only work in the z variable, soo the expression (6) can be rewritten as:
(7)
Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):
(S_b)