Figure shows, in cross section, three solid cylinders, each of length L and uniform charge Q. Concentric with each cylinder is a cylindrical Gaussian surface, with all three surfaces having the same radius. Rank the Gaussian surfaces according to the electric field at any point on the surface, greatest first.

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$II \to I \to III$

$I \to II \to III$

$III \to II \to I$

all tie

answer is D.

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Detailed Solution

$\begin{array}{l} In each case: ∮ \vec{E} \cdot d \vec{A} = \frac{q_{enclosed}}{ε_{0}} \\ \begin{aligned} E ∮ d A = \frac{Q}{ε_{0}} \\ E (4 π r^{2}) = \frac{Q}{ε_{0}} \\ E = \frac{1}{4 π ε_{0}} \frac{Q}{r^{2}} \end{aligned} \end{array}$

As charged enclosed within the Gaussian surface is the same, the radius of Gaussian surface is equal. Hence, the magnitude of electric field will be equal.

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