Question

If you put a total of 8.05×106×106 electrons on an intially electrically neutral wire of length 1.03 m, what is the magnitude of the electric field a perpendicular distance of 0.201 m away from the center of the wire?

Answer 1

The magnitude of the electric field is 0.1108 N/C

Step-by-step explanation:

Given;

number of electrons, e = 8.05 x 10⁶

length of the wire, L = 1.03 m

distance of the field from the center of the wire, r = 0.201 m

Charge of the electron;

Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)

Q = 1.2896 x 10⁻¹² C

Linear charge density;

λ = Q / L

λ = (1.2896 x 10⁻¹² C) / (1.03 m)

λ = 1.252 x 10⁻¹² C/m

The magnitude of electric field at r = 0.201 m;

`E = ((1)/(4 \pi \epsilon_o) )( 2 \lambda)/(r) \\\\E = k ( 2 \lambda)/(r)\\\\E = (8.89*10^9)*(2*1.252*10^(-12))/(0.201) \\\\E = 0.1108 \ N/C`

Therefore, the magnitude of the electric field is 0.1108 N/C