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Figure 1 depicts the cross section of a long, thin, uniformly charged, conducting cylinder. The radius $b$ is twice the length of radius $a$, which lies just beyond the outer surface of the cylinder.

What is the ratio $R = \cfrac{E(b)}{E(a)}$ of the electric field at radius $b$ to the field at radius $a$?

A

$R = 2$.

B

$R = \cfrac{1}{4}$.

C

$R = \cfrac{1}{2}$.

D

$R = 0$.

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