Question

A straight, nonconducting plastic wire 8.00cm long carries a charge density of 130nC/m distributed uniformly along its length. It is lying on a horizontal tabletop.

Find the magnitude and direction of the electric field this wire produces at a point 5.00cm directly above its midpoint

Answer 1

The magnitude of the electric field produced by the wire at a point 5.00cm above its midpoint is 146 kN/C and it points downwards due to the positive charge density of the wire.

Step-by-step explanation:

To find the magnitude and direction of the electric field produced by the wire at a point 5.00cm above its midpoint, we can use the formula for the electric field due to a uniformly charged line:

E = (λ / 4πε₀) * (1 / r)

where E is the electric field, λ is the charge density of the wire, ε₀ is the permittivity of free space, and r is the distance from the wire. In this case, λ = 130nC / 0.08m = 1.625μC/m and r = 0.05m. Plugging these values into the formula:

E = (1.625μC/m / 4π(8.85x10^-12 C²/Nm²)) * (1 / 0.05m) = 146 kN/C

The magnitude of the electric field is 146 kN/C, and it points downwards since the wire has a positive charge density.