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(A) The electric field $E$ due to an infinitely long straight wire with linear charge density $\lambda$ at a distance $r$ is given by the formula:$E =

Vedclass · Accepted Answer

$0.33 	imes 10^{11} \, N \cdot C^{-1}$

Vedclass · Answer

(A) The electric field $E$ due to an infinitely long straight wire with linear charge density $\lambda$ at a distance $r$ is given by the formula:$E = \frac{\lambda}{2 \pi \varepsilon_0 r}$This can be rewritten as:$E = \frac{2 \lambda}{4 \pi \varepsilon_0 r} = 2k \frac{\lambda}{r}$where $k = \frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 \, N \cdot m^2 \cdot C^{-2}$.Given:$\lambda = \frac{1}{3} \, C \cdot m^{-1}$$r = 18 \, cm = 0.18 \, m = 18 	imes 10^{-2} \, m$Substituting the values:$E = 2 	imes (9 	imes 10^9) 	imes \frac{1/3}{18 	imes 10^{-2}}$$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} \, N \cdot C^{-1}$$E \approx 0.33 	imes 10^{11} \, N \cdot C^{-1}$

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

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