# Can you miss out a factor in scalar potential?

1. Sep 3, 2013

### DunWorry

I've been wrestling with this for a few days (not literally). I got confused because I read in a book that E = - ∇ $\phi$ where E is the electric field and $\phi$ is the scalar potential. However in my notes I had that for a conservative force F = -∇$\phi$. I got confused because electric force and electric field are not the same thing, but I eventually realised that the $\phi$ in force is potential energy and not potential as it is with the electric field.

A long time ago I recall someone telling me that you could miss out a factor in scalar potential. Is this right? my reasoning was that because potential and potential energy only differ by a constant factor for example q (charge), and if you were dealing with just scalar potential and not potential energy you could remove this factor?

On the enclosed attatchment, they are showing that the line integral for work on a conservative field can be written as difference in potential. It looks like it should be = -3$\int d\phi$ but they just write = -$\int d\phi$, have they missed out the factor of 3?

I'm sorry if what I have said is complete BS, but I wanted to get it cleared up =)

Thanks

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2. Sep 3, 2013

### the_wolfman

These are two different concepts.

First if a force is conservative we can write it as $F= \nabla \psi$. Here $\psi$ is a generic potential field. It could be related to the electric potential but it does not have to be. For example it could be the gravitational potential. Sometimes the is an additional constant out front $F= k \nabla \phi$. It the case of the electric potential its the electron charge. However we can pull the constant inside the gradient and the initial statement still holds.

$F= k\nabla \phi = \nabla k\phi = \nabla \psi$

The second statement
has to do with the fact that the force is a derivative of the potential. Therefore you can add a constant $\phi_0$ to the potential and the force will remain unchanged.

$\nabla (\phi(x) + \phi_0) = \nabla \phi(x) + \nabla \phi_0 = \nabla \phi(x) + 0$

3. Sep 5, 2013

### DunWorry

Ok sorry I'm a little confused. $F= \nabla \psi$, where $\psi$ is the scalar potential field, or potential energy? I have seen both in different places, or are they the same thing? From my understanding. A scalar potential field is a field which gives you a scalar value at a certain point. Am I correct in thinking that the scalar value is interpreted as the potential at that point? But the potential energy is a constant times the potential eg. Mass for gravity and charge for electric potential energy? Could you please clarify a little more on what $\psi$ is?

Am I correct in thinking when its written $E= \nabla \psi$, where E is the electric field, the $\psi$ is the scalar potential field. This makes sense because Electric field is force per unit charge, and the scalar potenial is the potential energy per unit charge. When its written $F= - \nabla \psi$ and F is the electric force, not the electric field, the $\psi$ is the potential energy? and NOT potential? It makes sense because the two expressions are the same except you multiply one by the same constant?

I see so that has to do with the fact you can choose an arbitrary point to be zero potential? Also when I said miss out a factor in scalar potential, I meant a multiplying factor eg. 2x (on the attatched thumbnail it should be integral of 3 d(phi) right?)

Last edited: Sep 5, 2013
4. Sep 5, 2013

### Staff: Mentor

In this case, $\psi$ is the potential energy. It can get confusing because sometimes we talk about the force and sometimes we talk about the field that produces the force. Take the electric field and force, to be speciflc:

In terms of the electric field, $\vec E = -\vec \nabla V$, where V is the electric potential (volts) as a function of position.

In terms of the electric force on a certain charge, $\vec F = -\vec \nabla U$, where U is the electric potential energy (joules) of that charge, as a function of position.

The two equations are related by a factor of q on both sides: $\vec F = q \vec E$ and U = qV.

5. Sep 5, 2013

### the_wolfman

No, the factor of three is wrong. The problem is that the derivatives are partial derivatives of phi, not complete derivatives and $\frac {\partial \phi \left(x,y,z\right) }{\partial x}dx \neq d\phi$

In my example $\psi$ is a general scalar potential. I switched variables because I didn't want you to try to associate it with anything related to the electric field. It is just a generic potential.

The point I was trying to make is that any time you can take a force and write it in the form $F = \nabla \psi$ then we know that its conservative. And when we do so we call $\psi$ a potential field.

This is actually a math result, not a physics result

Recall that work is
$W = \int_a^b \vec F \cdot dl$
If we can write $F = \nabla \psi$ then the above expression simplifies
$W = \int_a^b \vec F \cdot dl = \int_a^b \nabla \phi \cdot dl = \phi(b) - \phi(a)$

The important part is that the work done in going from a to b does not depend on the path. And if we go from a to b and then back to a again the total work done will be 0.

Now writing $F = \nabla \psi$ is the way a mathematician might define a potential. For historical reasons and utility we often define physical potentials slightly differently. For instance we define the electric potential using $F_E = -q \nabla \phi$. The charge is constant, so it won't affect the above result dealing with work. In this case the mathematicians definition of potential is related to the electric potential $\psi = -q \phi$

6. Sep 5, 2013

### DunWorry

hmmm I understand what you say, do you mind explain this point a little further? sorry I know this is kind of a maths related question now. Thanks for the other posts and replies, I'm glad I got that cleared up =)