Question

A very long uniform line of charge with charge per unit length λ = +5.00 μC/m lies along the x-axis, with its midpoint at the origin. A very large uniform sheet of charge is parallel to the xy-plane; the center of the sheet is at z = +0.600 m. The sheet has charge per unit area σ = +8.00 μC/m², and the center of the sheet is at x = 0, y = 0. Point A is on the z-axis at z = +0.300 m, and point B is on the z-axis at z = -0.200 m. What is the potential difference Vₐᵦ = VA - Vᵦ between points A and B? Which point, A or B, is at higher potential?

Answer 1

The potential difference between points A and B is 0V. Point A is at a higher potential than point B.

Step-by-step explanation:

To find the potential difference between points A and B, we can use the formula V = -E * d, where V is the potential difference, E is the electric field, and d is the distance between the two points. Since point A is at a higher z-coordinate than point B, we can assume that the electric field is directed in the positive z-direction. Therefore, the potential at point A will be higher than the potential at point B.

First, we need to find the electric field. The electric field due to a uniformly charged line is given by E = (λ / (2πε₀)) * ln(r2/r1), where λ is the charge per unit length, ε₀ is the permittivity of free space, and r2 and r1 are the distances from the line charge to the point where the electric field is being measured. In this case, we need to find the electric field at the midpoint of the line charge, so r2 = r1 = 0.6 m.

Substituting the given values into the formula, we get E = (5.00 μC/m / (2π * 8.85 x 10^-12 C^2 / Nm^2)) * ln(0.6/0.6) = 0 N/C.

Since the electric field is 0, the potential difference V = -E * d = 0V between points A and B.