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Re: Sign of Work (electric)



I see nothing wrong with your description. Like Ken's
analogy, it should help to prevent the following error.

After writing dV=-E*ds ask this --> "Suppose that the
direction of the x axis (where points A and B are situated)
is reversed. Will the sign of dV change or not?" Many
students would probably answer "yes" to this question.
Ludwik Kowalski

David Ward wrote:

Dear Colleagues:

Ludwik and Ken write:

In trying to justify the negative sign in dV=-E*ds, where
E and ds are the electric field and the displacement, I will
emphasize the dot product nature of E*ds. The potential
energy (per unit charge) increases when the dot product
is negative (obtuse angle between E and ds), otherwise
it decreases, or remains the same. The "uphill" is locally
defined by the direction of E; it has nothing to do with
the orientation of the x,y,z axes (with respect to local E).
There is no ambiguity here, as long as we agree that the
probe charge is positive. Right?
Ludwik Kowalski

I think "downhill" ( lower V) is in the direction of E, just like
with g. So I have no problem with dV = -E *ds (* dot product)
Ken Fox

It's been a busy week, and I may have missed some of the thread, but
please allow me to comment. This topic was a great struggle for me
as an undergrad when I first encountered it.

"Uphill" and "downhill" are a bit vague. If we use those terms they
should refer to energy only, shouldn't they? I try to make a point to
my students that when released from rest objects move in such a way
as to lower their potential energy. After the students are well-versed
in mgh energy conservation problems, then a more general definition of
potential energy may be posed: "The change in potential energy between
two points is equal to the work an agent must do against the field to
move an object at fixed speed from point A to point B." It takes some
time to develop this statement, but I then show that it retrieves our old
mgh term. Based on the definition: dU = F(agent)*ds, where dU is a
change in the potential energy. But in opposing the field as stated in the

definition, F(agent)=-F(field), so we have dU=-F(field)*ds. When
we apply this to the electric field, we divide both sides by charge q
and get dV=-E*ds, with, of course, E=F(field)/q. So the minus
sign really comes from the definition, which leads us reflect with
our students that the definition will embody the idea that when released
from rest objects- be they masses in a gravitational field or charges in
an electric field- will move in such a way as to decrease their potential
energy.

Do you folks think my definition is okay, or does it need improvement?
Thanks folks! Have an excellent weekend!