Ratio of electric fields due to cylindrical charge of infinite length at a distance equal to its radius from its surface to that from its surface to that from its axis is

3

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\frac{1}{3}

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\frac{1}{2}

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Solution

The correct option is D $\frac{1}{2}$
$E_{1} = \frac{2 k λ}{2 R} = \frac{k λ}{R}$
$E_{2} = \frac{2 k λ}{2 R}$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{1}{2}$

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Ratio of electric fields due to cylindrical charge of infinite length at a distance equal to its radius from its surface to that from its surface to that from its axis is

The correct option is D 12E1=2kλ2R=kλR E2=2kλ2R ⇒E1E2=12

The correct option is D $\frac{1}{2}$
$E_{1} = \frac{2 k λ}{2 R} = \frac{k λ}{R}$
$E_{2} = \frac{2 k λ}{2 R}$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{1}{2}$