Topic open: Electric fields, potentials and capacitors

The following topic is now open for discussion: Electric fields, potentials, capacitors and parallel plates

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Electric fields and electrostatic potential

I've had an interesting discussion regarding Electric field and Electric potential and thought I would repost it here so it wasn't buried in the comments of an old post...

In relation to the post Question: E=V/d or V=-Ed? (to minus or not to minus), Peter asked:

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.



Is this way of thinking correct?

Thanks!

Which is exactly right. In fact, the "full" form (using vector calculus) is given by:



which essentially says to add up (integrate) the components of the electric field parallel to the path taken (the dl). If the electric field is constant and the path is straight, then everything reduces to what we have above.

Conversely, the electric field is proportional to a sort of directional slope of the potential (called the gradient) such that the greatest forces will be felt by charges in the "steepest" regions of electrostatic potential. More on that later perhaps...