Question

An infinite line of charge produces a field of magnitude$\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{C}$at distance 2.0 m. Find the linear charge density.

Short answer

The linear charge density is $5.0\times {10}^{-6}\mathrm{C}/\mathrm{m}$.

Step-by-step solution

The given data

1. Magnitude of an electric field,$\left|\mathrm{E}\right|=4.5\times {10}^4\mathrm{N}/\mathrm{C}$
2. Distance of the line from the electric field,$\mathrm{r}=2.0\mathrm{m}$

Understanding the concept of the electric field

Using the concept of the electric field of a cylindrical Gaussian surface, we can get the required value of the linear charge density by substituting the given data.

Formula:

The electric field of a cylindrical Gaussian surface,$\left|\mathrm{E}\right|=\frac{\lambda }{2\pi {\epsilon }_0\mathrm{r}}$ (1)

Calculation of the linear charge density

Substituting the given values in equation (1), we can get the required value of the uniform linear density on the line as follows:

$$\begin{gathered} \mathrm{\lambda }=2{\mathrm{\pi \epsilon }}_0\mathrm{rE} \\ =2\mathrm{\pi }\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)\left(4.5\times {10}^4\mathrm{N}/\mathrm{C}\right)\left(2.0\mathrm{m}\right) \\ =5.0\times {10}^{-6}\mathrm{C}/\mathrm{m} \end{gathered}$$

Hence, the value of the linear density is $5.0\times {10}^{-6}\mathrm{C}/\mathrm{m}$.