Work on a charge inside a Variable Electric Field

[Alan I]

Homework Statement

A variable electric field permeates all space:

E= (5.04x+ 84.2)x10⁴i where x is in meters and E is in N/C

How much work is done to move point charge q = 8.03 μC at constant velocity along the x-axis from point A at (3.63 m,0) to (7.67 m,0)? The sign will indicate who does the work.

NOTE: This requires integration!

Homework Equations

ΔU=-q∫Eds

The Attempt at a Solution

Δu=-(8.03x10^-6x10⁴∫(5.04X+84.2)dx
-(8.03x10^-2)*[5.04X²/2+84.2X]_3.63^7.67
=-30.62J

Are the limits wrong or the integral itself? Any hints would be greatly appreciated!

 

[Orodruin]

You have written the change in potential energy. If the change in potential energy is negative, the work on the charge is positive. Compare with computing the work as the line integral of the force.

 

[Alan I]

You have written the change in potential energy. If the change in potential energy is negative, the work on the charge is positive. Compare with computing the work as the line integral of the force.

OK I think I got it. So the mistake was the sign bc ΔU=-W=-∫F*dl , so since I got ΔU=-30J ⇒ W=30J.

 

[rude man]

I think the original answer is correct. The E field imparts a force in the +x direction on the charge so the work done ON the charge is negative.