# Negative and Positive E-Field Components

1. Feb 22, 2016

### Drakkith

Staff Emeritus
1. The problem statement, all variables and given/known data
In the diagram below, each division on the horizontal axis (the displacement s) is 0.10 m while each division on the vertical axis (the electric potential V) is 1.0 V. What is the electric field component along the displacement axis s in each of the regions specified below? Include the sign of the value in your answer.
(a) from 0.0 m to 0.10m
(b) from 0.10m to 0.90m
(c) from 0.90m to 1.00m

2. Relevant equations

ΔV=Vf-Vi = -∫E⋅ds

3. The attempt at a solution
I understand that the e-field for a and c is zero. A constant potential means that the e-field isn't changing.
My question is about b. The answer is -6.25 V, but I can't wrap my head around what this means. What's the difference between an e-field of +6.25 V and -6.25 V? Is it just the direction the e-field points compared to the direction you're moving?

2. Feb 22, 2016

### Staff: Mentor

That's about it. The field vector should respect the coordinate system that's in place. So here the spacial component is the s-axis, increasing to the right, and the potential increases to the right, too. So the e-field vector, pointing from higher potential to lower potential, points to the left.

3. Feb 22, 2016

### Drakkith

Staff Emeritus
Alright. So then the e-field would perform negative work on a positive test charge brought in from the left, and therefor the change in potential, the negative of the work done divided by the charge, is positive, right?

4. Feb 22, 2016

### Staff: Mentor

Right. I sometimes find it helpful to imagine that the region in question is bordered by two plates, like in a capacitor. The potential difference between the plates is established by a voltage supply of the appropriate size. Then I can easily picture the field lines and what a charged particle situated in or moving through the region will experience.