An infinitely long thin straight wire has uniform linear charge density of 1/3 coul. $m^{- 1}$ . Then the magnitude of the electric intensity at a point 18 cm away is

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$3 \times 10^{11} {NC}^{- 1}$

$1.32 \times 10^{11} {NC}^{- 1}$

$0.66 \times 10^{11} {NC}^{- 1}$

$0.33 \times 10^{11} {NC}^{- 1}$

answer is A.

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Detailed Solution

Given Data: For an infinitely long wire, $λ = 1 / 3 c o u l . m^{- 1}, r = 18 cm$

Formula Used: $E = \frac{2 k λ}{r}$

Detailed Solution:

The field due to an infinite linear charge density is,

$E = \frac{2 k λ}{r}$

$= \frac{2 \times 9 \times 10^{9} \times 1 / 3}{18 \times 10^{- 2}}$

$= 0.33 \times 10^{11} {NC}^{- 1}$

Hence, the correct option is (A).

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