# An infinitely long thin wire carries a uniform charge per unit length q. Find the electric field at a distance of R units from the wire?

Jan 29, 2017

Electric field is $\frac{q}{4 {\epsilon}_{0} R}$, where $q$ is the uniform charge per unit length and ${\epsilon}_{0}$ is electrical permittivity of free space

#### Explanation:

Consider the infinitely long thin wire to carry a uniform charge per unit length $q$, The field created by a small portion $\mathrm{dx}$ of it at a distance $R$ will be

$\frac{1}{4 \pi {\epsilon}_{0}} \times \frac{q \mathrm{dx}}{\sqrt{{x}^{2} + {R}^{2}}}$, where ${\epsilon}_{0}$ is electrical permittivity of free space

and total field created by thin wire will be

$\frac{1}{4 \pi {\epsilon}_{0}} {\int}_{- \infty}^{\infty} \frac{q \mathrm{dx}}{\sqrt{{x}^{2} + {R}^{2}}}$

= $\frac{q}{4 \pi {\epsilon}_{0}} {\int}_{- \infty}^{\infty} \frac{\mathrm{dx}}{\sqrt{{x}^{2} + {R}^{2}}}$

= $\frac{q}{4 \pi {\epsilon}_{0}} \frac{1}{R} | {\tan}^{- 1} \left(\frac{x}{R}\right) {|}_{- \infty}^{\infty}$

= $\frac{q}{4 \pi {\epsilon}_{0} R} \left(\frac{\pi}{2} - \left(- \frac{\pi}{2}\right)\right)$

= $\frac{q}{4 {\epsilon}_{0} R}$