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Electric field, Electric Potencial, Electric Force, Potential Electric Energy 
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#1
Feb2310, 03:27 PM

P: 109

Hi, I have some questions about these concepts.
The equations: [tex]Electric field =\frac{K.Q}{d^2}[/tex] [tex]Electric potential =\frac{K.Q}{d}[/tex] [tex]Electric Force =\frac{K.Q.q}{d^2}[/tex] [tex]Potential Electric Energy =\frac{K.Q.q}{d}[/tex] Electric field and electric force are vectors, electric potential and potential electric energy are scalar values. Note that some of the concepts are related with each other just adding *d on the equation. For example electric field and electric potential are almost the same equation except to the fact that electric field is inverse proportional to the d squared and electric potential is inverse proportional only to the d. Also electric field is vector and electric potential is scalar. How can the electric field be a vector when its only the electric potential (scalar) multiplied by d? What's the intuitive meaning of the relation between electric field and electric potential (as the relation of electric force and potential electric energy)? These four concepts are really necessary to fully describe a charge and the space that it's contained? Or some of them were created just to make the calculations easy? I also realized the following: if you have a charge alone in a space you don't have potential electric energy because you don't have other charge to cause force on it, right? So, if I add a charge to the space, these two charges acquires potential electric energy and they start approximating (considering they have opposite signs). But, when distance between them decrease, the potential electric energy increases, right? So, what's the meaning of that? I'm confused... I would be grateful if someone help me understanding these concepts. Thank you 


#2
Feb2310, 06:46 PM

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P: 5,196

field  E potential  V force  F potential energy  U [tex] \textbf{E} =  \nabla V [/tex] The gradient operator, symbolized by [itex] \nabla [/itex] is like a derivative in more than one dimension. If you haven't taken calculus and hence haven't done derivatives, don't worry about it. Just think of it this way: the electric potential has a scalar value at every point in space. The electric field is a corresponding vector at every point space whose magnitude tells you the maximum rate of change (with position) of the electric potential, and whose direction indicates the direction in which that rate of change is calculated. So, the electric field points in the direction in which the rate of change of electric potential (in space) is a maximum. If you have taken calculus, then note that for the simple 1D case represented by coordinate 'r', or a case with spherical symmetry, the equation I posted above would reduce to: [tex] \textbf{E} =  \frac{\partial V}{\partial r} \hat{\textbf{r}} [/tex] [tex] \textbf{E} = \frac{kQ}{r^2}\hat{\textbf{r}} [/tex] Where the r with a hat on top of it is a unit vector that always points radially outward from the source. If you compute E by taking the gradient of V, then this additional unit vector is already taken care of, because the gradient is a vector operator: it includes the unit vector as part of it. Likewise, the relationship between force and electric potential energy is: [tex] \textbf{F} =  \nabla U [/tex] 


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