There is a rod that is 50cm long and has a radius of 1cm that carries a charge of 2uC distributed uniformly over its length. What would be the value of the magnitude of the electric field. (a) 4.0 mm from the rod surface, not either end

So far I have 2(8*10^-6)(9*10^9)/.55=261818/1.4= 1.9*10^6

And what do these numbers represent? Did you read my prior post? http://www.jiskha.com/display.cgi?id=1251172915

To calculate the magnitude of the electric field at a point 4.0 mm from the surface of the rod, you need to use the equation for the electric field due to a uniformly charged rod.

The equation for the electric field E at a point on the axial line of a uniformly charged rod is given by:

E = (k * λ) / (2πε * r)

Where:
k = Coulomb's constant (9 x 10^9 Nm^2/C^2)
λ = charge per unit length of the rod (2 x 10^-6 C/0.5 m = 4 x 10^-6 C/m)
r = distance from the surface of the rod (0.04 m in this case)
ε = permittivity of free space (8.85 x 10^-12 C^2/Nm^2)

Substituting the values into the equation:

E = (9 x 10^9 Nm^2/C^2 * 4 x 10^-6 C/m) / (2π * 8.85 x 10^-12 C^2/Nm^2 * 0.04 m)

Simplifying:

E = (36 x 10^3) / (2π * 8.85 x 10^-12 * 0.04)

E = 127446.81 N/C

So, the magnitude of the electric field at a point 4.0 mm from the surface of the rod, not at either end, is approximately 127446.81 N/C.