High School

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What is the relationship between electric field (E) and radius (r) for a cylindrical Gaussian shell?

A. [tex]E \propto r[/tex]
B. [tex]E \propto \frac{1}{r}[/tex]
C. [tex]E \propto r^2[/tex]
D. [tex]E \propto \frac{1}{r^2}[/tex]

Answer :

Final answer:

The electric field (E) is inversely proportional to the radius (r) for points outside a cylindrical Gaussian shell, following the relationship given by Gauss' Law, which states that E is proportional to 1/r for a cylindrical charge distribution. The correct option is b.

Explanation:

The relationship between the electric field (E) and radius (r) for a cylindrical Gaussian surface is described by Gauss' Law, which relates the electric flux through a closed surface to the charge enclosed by that surface. When considering a cylindrical charge distribution, such as a uniformly charged long cylinder, the electric field outside the cylinder (for points where r > R, the radius of the cylinder) is inversely proportional to the radius. Specifically, E is proportional to 1/r. Inside the cylinder (for points where r < R), the electric field increases linearly with distance from the central axis because the amount of charge enclosed by the Gaussian surface increases as the radius increases.

Therefore, the correct option for the relationship between E and r for points outside a cylindrical Gaussian shell is b) E
Proportional 1/r.

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