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(A) Given:Linear charge density,$\lambda = \frac{1}{3} ~Cm^{-1}$Distance,$r = 18 ~cm = 18 	imes 10^{-2} ~m$Permittivity of free space,$\varepsilon_0

Vedclass · Accepted Answer

$0.33 	imes 10^{11} ~NC^{-1}$

Vedclass · Answer

(A) Given:Linear charge density,$\lambda = \frac{1}{3} ~Cm^{-1}$Distance,$r = 18 ~cm = 18 	imes 10^{-2} ~m$Permittivity of free space,$\varepsilon_0 \approx 8.85 	imes 10^{-12} ~C^2 N^{-1} m^{-2}$The electric field intensity $E$ at a distance $r$ from an infinitely long thin straight wire is given by the formula:$E = \frac{\lambda}{2 \pi \varepsilon_0 r}$We can rewrite this as:$E = \frac{2 \lambda}{4 \pi \varepsilon_0 r} = 2k \frac{\lambda}{r}$where $k = \frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 ~Nm^2 C^{-2}$.Substituting the values:$E = 2 	imes (9 	imes 10^9) 	imes \frac{(1/3)}{18 	imes 10^{-2}}$$E = 18 	imes 10^9 	imes \frac{1}{3 	imes 18 	imes 10^{-2}}$$E = \frac{18 	imes 10^9}{54 	imes 10^{-2}}$$E = \frac{1}{3} 	imes 10^{11} ~NC^{-1}$$E \approx 0.33 	imes 10^{11} ~NC^{-1}$

An infinitely long thin straight wire has a uniform linear charge density of $\frac{1}{3} ~Cm^{-1}$. The magnitude of the electric field intensity at a point $18 ~cm$ away is (given $\varepsilon_0 = 8.85 \times 10^{-12} ~C^2 N^{-1} m^{-2}$):

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