Negative and Positive E-Field Components

[Drakkith]

Homework Statement

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

Homework Equations

ΔV=V_f-V_i = -∫E⋅ds

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?

 

[gneill]

That's about it. The field vector should respect the coordinate system that's in place. So here the spatial 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.

 

[Drakkith]

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?

 

[gneill]

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?

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.