# (Electric Charge and Electric Field)

Consider point a which is 70 cm north of a -3.6 µC point charge, and point b which is 84 cm west of the charge (Fig. 17-23).

(a) Determine Vba = Vb - Va.
V = ____
(b) Determine Eb - Ea.
Magnitude
N/C = ____
Direction = _____
° (counterclockwise from east is positive)

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At first I was thinking the first part had to do with the E = (KQ1)/r2 equation. You take that for both point a and b then subtract b-a to get the final velocity. That didnt seem to work, and I have been looking frantically for the past 30 min through my notes and have not been able to find an equation to start off this problem. I think I may be confusing test charges with normal charges.

dextercioby
Homework Helper
That is not velocity,that is ELECTRIC POTENTIAL.And you shouldn't be using the formula for the electric field,but the formula for the electric potential...

Daniel.

Hmmm.... tried the equation V=Ed for potential but that didnt work. Am I using the right equation?

dextercioby
Homework Helper
No,u should be using the electric potential created by an electric charge...

$$V(r)=\frac{1}{4\pi\epsilon_{0}\epsilon_{r}} \frac{q}{r}$$

Daniel.

Haven't had a chance to use that equation... can you explain where the e sub 0 and r come from please?

dextercioby
Homework Helper
Those are electric permitivities.The one with the subscript "0" is the electric permitivity of vacuum.The one with the "r" subscript is the relative electric permitivity of the medium wrt the vacuum and is dimensionless...

Irrelevant into discussion,really.U got a problem to solve...

Daniel.

I don't think it's irrelevant, I like to know all the values and what they are if I am going to use an equation. I just find it hard to believe, Axeman, you dont know what $$\epsilon_{0}$$ is since it seems you are in an electricity class.

dextercioby
What do you mean...??He SHOULD KNOW what $\epsilon_{0}$ is.Even if he's in High School...