Potential energy calculation of charges in an electric field?

You find the voltage change from one location to the next and then multiply that by the charge. Let's do it:

6200V/m * 0.08m = 496V = 496J/C

The voltage DECREASES as you move further to the right on the x axis. How do I know that? If you let the positive charge go, it would fly to the right, it's kinetic energy increasing and potential energy decreasing.

496J/C * 16e-6C = 7.936e-3 J <--------

Moving the charge anywhere in the y axis direction will not change it's PE in a uniform electric field in the x direction so the second answer = 0J

(1)

Filed E = 6200 V/m

Change is potential = (-6200 V/m) (8/100 m) ... // (E = ∆V/∆r => ∆V = -E ∆r)

= -496 V (or J/C)

Change Potential Energy:= q ∆V = (16 x 10⁻⁶ C) (-496 J/C)

= -7.936 x 10⁻³ J or -7.936 mJ

Potential energy will fall by 7.936 mJ

(2)

We have seen

∆V = -E ∆r

In vector form

∆V = -E̅ ∙ ∆r̅ = 0 (Since direction of change of ∆r̅ and that of E̅ are perpendicular to each other.)

Change is potential energy = 0