An infinitely long thin straight wire has uniform linear charge density of $\frac{1}{3} \mathrm{Cm}^{-1}$. Then, the magnitude of the electric intensity at a point $18 \mathrm{~cm}$ away is (given $\varepsilon_0=8.8 imes 10^{-12} \mathrm{C}^2 \mathrm{Nm}^{-2}$ )

Question

An infinitely long thin straight wire has uniform linear charge density of $\frac{1}{3} \mathrm{Cm}^{-1}$. Then, the magnitude of the electric intensity at a point $18 \mathrm{~cm}$ away is (given $\varepsilon_0=8.8 	imes 10^{-12} \mathrm{C}^2 \mathrm{Nm}^{-2}$ )

Aicharya · Accepted Answer

Charge density of long wire $\begin{aligned} \lambda & =\frac{1}{3} \mathrm{C}-\mathrm{m} \\ \text { and } \quad r & =18 \times 10^{-2} \mathrm{~m} \end{aligned}$

From Gauss theorem $\begin{gathered} \oint \overrightarrow{\mathbf{E}} d \overrightarrow{\mathbf{S}}=\frac{q}{\varepsilon_0} \\ E \oint d S=\frac{q}{\varepsilon_0} \end{gathered}$ or $\quad E \times 2 \pi r l=\frac{q}{\varepsilon_0}$ or $\quad E=\frac{q}{2 \pi \varepsilon_0 r l}=\frac{q / l}{2 \pi \varepsilon_0 r}$ $\begin{aligned} & =\frac{\lambda \times 2}{2 \pi \varepsilon_0 r \times 2}=\frac{\lambda \times 2}{4 \pi \varepsilon_0 r} \\ & =9 \times 10^9 \times \frac{1}{3} \times 2 \times \frac{1}{18 \times 10^{-2}} \\ & =\frac{1}{3} \times 10^{11}=0.33 \times 10^{11} \\ & =0.33 \times 10^{11} \mathrm{NC}^{-1}\end{aligned}$

An infinitely long thin straight wire has uniform linear charge density of $\frac{1}{3} \mathrm{Cm}^{-1}$.…

Solution