Can E=V/d be used to calculate the electric field of a dipole?

[LostThoughts]

This is from a lab where my team found the equipotential lines from the electric field of a dipole. The information I was given to calculate the magnitude of the electric field seems too simple, and for some reason I'm expecting this to be more complex. So, this is my attempt at double checking. Can E=V/d be used for all of this?

1. Homework Statement
Determine the magnitude of the average electric field between two sets equipotential lines.
0.5v to 1v d=0.022m
5v to 5.5v d=0.022m

Determine the magnitude of the average electric field between the two pins.
pinnegative=0v pinpositive=6v d=0.087m

Determine the magnitude of the average electric field between one pin and each equipotential line.
V=2v d=0.035m
V=3v d=0.042m
V=4v d=0.057m
V=5v d=0.069m

Homework Equations

E=F/q V=W/q W=Fd E=V/d

 

[kuruman]

Can E=V/d be used for all of this?

This is correct only if the field is uniform. The electric field generated by a dipole is not uniform. If you measure the potential difference ΔV in a direction perpendicular to a given equipotential, the electric field at that general point is approximately given by E = ΔV/Δx, where Δx is the distance over which you measured the potential difference. The smaller you make Δx, the more accurate your value for E. In your case you probably used a two-prong connector with about 1 cm separation between prongs, so you have no control over that.