Answer:[tex]E=59.33\times 10^6 N/C[/tex]
Explanation:
Given
linear charge density[tex](\lambda )=99 \mu C/m[/tex]
distance(r)=3 cm
Permittivity For free space [tex]\epsilon _0=8.85\times 10^{-12} \frac{C^2}{N.m^2}[/tex]
Charge enclosed by gaussian surface
[tex]q_{encl}=\lambda \times l[/tex]
the flux through gaussian surface
[tex]\phi =\oint EdA=E\left ( 2\pi rl\right )[/tex]
Also from Gauss law
[tex]\phi =\frac{q_{encl}}{\epsilon _0}[/tex]
Equate
[tex]E\left ( 2\pi rl\right )=\frac{q_{encl}}{\epsilon _0}[/tex]
[tex]E=\frac{\lambda }{2\pi r\epsilon _0}[/tex]
[tex]E=59.33\times 10^6 N/C[/tex]
