# Homework Help: Work on a charge inside a Variable Electric Field

1. Sep 16, 2015

### Alan I

1. The problem statement, all variables and given/known data
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!

2. Relevant equations
ΔU=-q∫Eds

3. The attempt at a solution
Δu=-(8.03x10-6x104∫(5.04X+84.2)dx
-(8.03x10-2)*[5.04X2/2+84.2X]3.637.67
=-30.62J

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

2. Sep 16, 2015

### Orodruin

Staff Emeritus
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.

3. Sep 16, 2015

### Alan I

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.

4. Sep 17, 2015

### 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.