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Can you miss out a factor in scalar potential? 
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#1
Sep313, 09:39 AM

P: 40

I've been wrestling with this for a few days (not literally). I got confused because I read in a book that E =  ∇ [itex]\phi[/itex] where E is the electric field and [itex]\phi[/itex] is the scalar potential. However in my notes I had that for a conservative force F = ∇[itex]\phi[/itex]. I got confused because electric force and electric field are not the same thing, but I eventually realised that the [itex]\phi[/itex] 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[itex]\int d\phi[/itex] but they just write = [itex]\int d\phi[/itex], 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 


#2
Sep313, 10:03 AM

P: 158

These are two different concepts.
First if a force is conservative we can write it as [itex]F= \nabla \psi [/itex]. Here [itex]\psi [/itex] 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 [itex]F= k \nabla \phi [/itex]. 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. [itex]F= k\nabla \phi = \nabla k\phi = \nabla \psi [/itex] The second statement [itex] \nabla (\phi(x) + \phi_0) = \nabla \phi(x) + \nabla \phi_0 = \nabla \phi(x) + 0 [/itex] 


#3
Sep513, 07:56 AM

P: 40

Am I correct in thinking when its written [itex]E= \nabla \psi [/itex], where E is the electric field, the [itex]\psi [/itex] 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 [itex]F=  \nabla \psi [/itex] and F is the electric force, not the electric field, the [itex]\psi [/itex] 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? 


#4
Sep513, 09:19 AM

Mentor
P: 11,778

Can you miss out a factor in scalar potential?
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
Sep513, 11:19 AM

P: 158

The point I was trying to make is that any time you can take a force and write it in the form [itex]F = \nabla \psi [/itex] then we know that its conservative. And when we do so we call [itex]\psi [/itex] a potential field. This is actually a math result, not a physics result Recall that work is [itex]W = \int_a^b \vec F \cdot dl [/itex] If we can write [itex]F = \nabla \psi [/itex] then the above expression simplifies [itex]W = \int_a^b \vec F \cdot dl = \int_a^b \nabla \phi \cdot dl = \phi(b)  \phi(a) [/itex] 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 [itex]F = \nabla \psi [/itex] 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 [itex]F_E = q \nabla \phi [/itex]. 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 [itex]\psi = q \phi [/itex] 


#6
Sep513, 04:59 PM

P: 40




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