Answer:
The value is [tex]E = 14.12 \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the cylinder is [tex]r = 12 \ cm = 0.12 \ m[/tex]
The density of the charge is [tex]\rho = 5.0 \ nC/m^3 = 5.0*10^{-9} \ C/m^3[/tex]
The position consider is a = 5.0 cm = 0.05 m
Gnerally from the magnitude of the magnetic field is mathematically represented as
[tex]E = \frac{\rho * s}{ 2 * \epsilon_o }[/tex]
Here [tex]\epsilon_o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85 *10^{-12} \ C/(V \cdot m)[/tex]
=> [tex]E = \frac{5.0*10^{-9} * 0.05}{ 2 * 8.85*10^{-12} }[/tex]
=> [tex]E = 14.12 \ N/C[/tex]
