Find the change in electric potential

[Brit412]

Homework Statement

I'm having trouble understand why I'm getting the following question wrong: a uniform electric field of magnitude of 6.0*10^5 N/C points in the positive x direction. Part A) Find the change in electric potential between the origin and the point (0, 6.0m)

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Homework Equations

Wouldn't it just be ∆V = -E*∆s = (6.0*10^5) (6.0m)?.

The Attempt at a Solution

 

[LowlyPion]

If the field is uniform in x direction and you want the voltage change from 0,0 to 0,6 it's 0. anything in the Y direction will be equipotential.

 

[nlightner]

Your solution is correct. The change in electric potential, ∆V, is equal to the negative of the electric field strength, E, multiplied by the displacement, ∆s. In this case, the electric field is uniform and points in the positive x direction, so the electric field strength is constant at 6.0*10^5 N/C. The displacement, ∆s, is the distance between the origin and the point (0,6.0m), which is simply 6.0m. Therefore, the change in electric potential is ∆V = -(6.0*10^5 N/C)(6.0m) = -3.6*10^6 J/C. This means that the electric potential decreases by 3.6*10^6 J/C as you move from the origin to the point (0,6.0m) in the direction of the electric field.