I was wondering why the notes from class say that for uniform Electric fields E=V/d, while the textbook and WebTA say that V=-Ed. Why is there no negative sign in the one case, but the negative is included in the other?
That is a very good question. I can first tell you that the reason my formula and the formula in the book are the same is that I copied mine out of the book. ;) I can also say that Prof. Altounian's notes are correct.
To get to the heart of this matter, really we have to think about what V is and what E is. The electric potential, V, is a scalar quantity. That is, it has a magnitude at a particular point in space, but no direction associated with it. But the electric field, E, is a vector quantity. This means it has both a magnitude and a direction. So if we look at a simple formula for the electric field, like E=V/d, we have to think this isn't the full story... how do we relate a vector and a scalar?? where is the direction part of E??? Well, it isn't there. Really all we get is the magnitude of E. So, whether you stick a negative sign there or not, you have to determine the direction of the electric field in another way (by determing where a high potential is and where a low potential is).
So, why did the text book (and myself) bother putting a negative sign there? Well, it comes from the calculus relation between E and V. If you have a map of V over an area, the electric field points downhill, so when you look at it from within a differential formalism you need the negative sign (so I'm used to seeing it that way).
In any case, I mean to change my webTA pages to agree with the class notes on this point. And the moral of the story is that since E is a vector you have to determine the direction as well as the magnitude.
10 comments:
Hey, I was just reading Cutnell & Johnson Physics 7th edition (Competitor textbook to Giancoli) and on pg 585 it had some comments about this. I am not sure if it is applicable?
They first derived the -form of the equation. So they said
W = Fd =qEd
But W = -PE
qEd = -PE
qEd/q = -PE/q
Ed = -V
V = -Ed
Ok so far so good.
then they had some comments about the form of the equation where V = Ed (no negative sign)
"When applied strictly to a parallel plate capacitor, however, this expression is often used in a slight different form. In figure 19.16, the metal plates of the capacitor are marked A( higher potential) and B (lower potential). Traditionally, in discussions of such a capacitor, the potential difference between the plates is referred to by using the symbol V to denote the amount by which the higher potential exceeds the lower potential. V = Va-Vb.
Thus,
E = -V/d = - (Vb-Va)/d = (Va-Vb)/d = V/d.
Would this be a valid explanation as well since in this course we are mostly dealing with parallel plate capacitors? Thanks.
Hi Peter,
Excellent idea to check out another text book, sometimes a different perspective is all you need...
However, I disagree with Cutnell's approach here. They are essentially taking advantage of two negatives which cancel, and I think it is confusing. It is true that, as a sort of shorthand, the potential difference between the plates is simply given as the amount by which the higher potential exceeds the lower. The problem I see with using this "no minus" version of the equation is that it does not represent the true relationship between E and V (ie. that the electric field points from the high potential to the low potential). This takes a relationship and turns it into an equation, which I dislike. Equations are limited in scope and easy to misapply, relationships can give you better understanding which is a much more solid basis.
I'd rather see you sketch a diagram showing the high and low potential and where the positive and negative charges are separated on the two plates and applying the "V" given as the magnitude of the potential difference. It will be much harder to go wrong with this picture in front of you, and the whole "-" issue more or less disappears.
I hope that helps.
Gee you are totally right. I posted this early this morning and now I already have a different idea, which you have already hinted here but I want to make sure I got this down 100%.
This just came to me.
Link to pic with equations and such. Blogger won't let me embed images in comments.
Is this way of thinking correct?
Thanks!
Peter, that's exactly it. :)
Do you mind if I post your explanation? I'd like to put our discussion in a post and add a brief note about how this works with vector calculus, because it ties in nicely with this discussion and I think it is not that difficult conceptually, and somewhat enlightening. Let me know if you are ok with this.
Sure, I am ok with it. Thanks for the help!
Hi my name is Abdulrahman. I from Egypt. I realy want to thank both of you because I was confused about this " - " case but now its very clear to me.
Thanks
Thanks! This helped me a lot too :)
Oh thank you all! It really really helped me clear this topsy-turvyness.
I love it when people doubt me. It makes me work harder to prove them wrong. See the link below for more info.
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