Question

An infinitely long solid cylindrical insulator of radius 12.0 cm has a non-uniform volume charge density of rho =4r³ where rho is in C/ m³ when r is in meters. Calculate the magnitude of the elect0072ic field at a distance of 22.00 cm from the axis of the cylinder.

Answer 1

The magnitude of the electric field at a distance of 22.00 cm from the axis of the cylindrical insulator is 3.47 * 10^8 N/C.

Step-by-step explanation:

To calculate the electric field at a distance of 22.00 cm from the axis of the cylindrical insulator, we can use the formula for the electric field due to a charged cylinder:

E = (rho * r)/(2 * epsilon_0),

where rho is the charge density, r is the distance from the axis, and epsilon_0 is the permittivity of free space.

Substituting the given values, we have rho = 4r³ C/m³ and r = 22.00 cm = 0.22 m. Plugging these values into the formula, we have:

E = (4(0.22^3))/(2 * epsilon_0) = (4(0.22^3))/(2(8.85 * 10^-12 N^-1 m^-2 C^2)) = 3.47 * 10^8 N/C.

Therefore, the magnitude of the electric field at a distance of 22.00 cm from the axis of the cylindrical insulator is 3.47 * 10^8 N/C.