Answer:
[tex]E = 9.4*10^{10}N/C[/tex]
Explanation:
We use Gauss's law which says
[tex]$\int E\cdot dA = \frac{Q_{enc}}{\varepsilon}. $[/tex] [tex](1)[/tex]
Now, for the cylindrical charge distribution the charge enclosed is
[tex]Q_{enc} = \rho V[/tex]
where [tex]V[/tex] is the volume of the cylinder.
To evaluate Gauss's law, the Gaussian surface we choose is a cylinder concentric with the charged cylinder; therefore, equation [tex](1)[/tex] becomes
[tex]E (2\pi rL )=\dfrac{\rho V}{\varepsilon _o}[/tex]
[tex]E (2\pi rL )=\dfrac{\rho \pi R^2L}{\varepsilon _o}[/tex]
[tex]E =\dfrac{\rho \pi R^2L}{ (2\pi rL )\varepsilon _o }[/tex]
[tex]\boxed{E =\dfrac{\rho R^2}{ 2\varepsilon _o r }}[/tex]
Putting in numerical values
[tex]\rho = 15C/m^3[/tex]
[tex]R = 17cm =0.17m[/tex]
[tex]r = 26cm=0.26m[/tex]
[tex]\varepsilon_0 =8.85*10^{-12}m^{-3}kg^{-1}s^4A^2}[/tex]
we get:
[tex]E =\dfrac{15 (0.17)^2}{ 2(8.85*10^{-12}) (0.26) }[/tex]
[tex]\boxed{E = 9.4*10^{10}N/C}[/tex]
