state gauss's theorem in electrostatics.apply this theorem to derive an expression for electric field intensity at a point near an infinitely long straight charged wire.

Dear Student,

Please find below the solution to the asked query:

Amount of Electric flux passing through a closed surface or the surface integral of Electric field intensity is equal to the ratio of the Charge enclosed by the surface and the Permittivity of the free space.

$\varphi =\frac{Q_{enclosed}}{{\epsilon }_0} \Rightarrow \int E.dS=\frac{Q_{enclosed}}{{\epsilon }_0}$

Electric Field due to a Long Straight Charged Wire:

Consider a long straight wire charge by a linear charge density λ as shown in the figure below.



To find the electric field at a distance from the wire, consider a gaussian surface as shown in the figure.

Let the electric field at the surface is E. Then,

From the definition of gauss law,

$$\begin{gathered} \int E.dS=\frac{Q_{enclosed}}{{\epsilon }_0}\Rightarrow E\int dS=\frac{Q}{{\epsilon }_0} \\ \Rightarrow E\left(2\pi RL\right)=\frac{Q}{{\epsilon }_0} \Rightarrow E=\frac{1}{2\pi {\epsilon }_{0}R}\left(\frac{Q}{L}\right) \\ \Rightarrow E=\frac{\lambda }{2\pi {\epsilon }_{0}R} \end{gathered}$$

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