Question

A very long, thin wire has a uniform linear charge density of 49 µC/m. What is the electric field (in N/C) at a distance 6.0 cm from the wire? (Enter the magnitude.)

Answer 1

1.467 x 10⁷ N/C

Step-by-step explanation:

Using Gauss's law, the electric field (E) on a charge enclosed by a gaussian surface is given by;

E = λ / (2π r ε₀) ---------------------(i)

Where;

λ = charge density in C/m

r = distance of the electric field from the wire in meters

ε₀ = permittivity of free space = 8.85 x 10⁻¹² Nm²/C²

From the question;

λ = 49 μC/m = 49 x 10⁻⁶C/m

r = 6.0cm = 0.06m

Substitute these values into equation (i) as follows, taking π = 3.142;

E = 49 x 10⁻⁶ / (2 x 3.142 x 0.06 x 8.85 x 10⁻¹²)

E = 49 x 10⁻⁶ / (3.34 x 10⁻¹²)

E = 14.67 x 10⁶ N/C

E = 1.467 x 10⁷ N/C

Therefore, the electric field at that distance is 1.467 x 10⁷ N/C