Answer:
Electric field, [tex]E=4.43\times 10^{10}\ N/C[/tex]
Explanation:
It is given that,
Radius of the cylinder, R = 12 cm = 0.12 m
Charge density, [tex]\rho=18\ C/m^3[/tex]
Let E is the electric field at a distance of r = 33 cm or 0.33 m from the axis. The relationship between the linear charge density and the volume charge density is given by :
[tex]\lambda=\rho\times \pi r^2[/tex]
Now according to Gauss's law :
[tex]E.2\pi rl=\dfrac{\lambda l}{\epsilon_o}[/tex]
[tex]E=\dfrac{\lambda}{2\pi \epsilon_or}[/tex]
or
[tex]E=\dfrac{\rho R^2}{2\epsilon_o r}[/tex]
[tex]E=\dfrac{18\times (0.12)^2}{2\times 8.85\times 10^{-12}\times 0.33}[/tex]
[tex]E=4.43\times 10^{10}\ N/C[/tex]
Hence, this is the required solution.
