Question

A uniform electric field, with a magnitude of 500 v/m, is points in the x direction. if the potential at x = 5.0 m is 2500 v, what is the potential at x = 2.0 m?

Answer 1

To find the potential at x = 2.0 m in a uniform electric field, we calculate the potential difference due to the change in position along the field direction and subtract it from the given potential at x = 5.0 m. The potential at x = 2.0 m is 1000 V.

Step-by-step explanation:

The student is asking about the potential at a different position in a uniform electric field. Given the electric field magnitude of 500 V/m, which points in the x direction, and the potential at x = 5.0 m being 2500 V, we can calculate the potential at x = 2.0 m. We know that the potential difference (V) in an electric field is given by the formula:

V = Ed,

where E is the electric field strength and d is the distance. To find the potential difference between x = 5.0 m and x = 2.0 m, we can multiply the electric field strength by the difference in distance:

ΔV = E Δx = 500 V/m × (5.0 m - 2.0 m),

ΔV = 500 V/m × 3.0 m = 1500 V,

Since the electric field points in the positive x direction, the potential decreases as we move in the direction of the field. The potential at x = 2.0 m will be:

V at x = 2.0 m = V at x = 5.0 m - ΔV,

V at x = 2.0 m = 2500 V - 1500 V = 1000 V.

Answer 2

V = E x

V = E * 5 = 2500 where E is defined at x = zero

E = 500 V / m

V(2.0) = 500 V/m * 2 m = 1000 Volts