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Electric field, Electric Potencial, Electric Force, Potential Electric Energy

  1. Feb 23, 2010 #1
    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. jcsd
  3. Feb 23, 2010 #2


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    I'll use the following symbols in the discussion that follows (boldface denotes vector quantities). Electric:

    field - E
    potential - V

    force - F
    potential energy - U


    Well, yes, but that's because the true mathematical relationship between electric field and electric potential is that the electric field is the negative gradient of the electric potential:

    [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] ​

    Your equation is for the magnitude of the electric field only. The full equation for the electric field should read:

    [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] ​

    I hope that my description above has made that clear. The electric force is the gradient of the electric potential energy. Hence, the force points in the direction of maximum rate of decrease in potential energy (this is true for gravitational forces and fields/energies as well). The charges will want to move in the direction that reduces the potential energy of the system.

    Yes, you are exactly right! Electric field and electric potential describe how a single charge influences its surroundings. But, if there is nothing in the surroundings to influence, then there will be no force and no corresponding potential energy. Potential energy is a property of a system of charges. It comes into play when you have more than one charge interacting through the electric force. The two quantities, F and U, describe this interaction.

    I don't understand this use of the word "approximating," it makes no sense. Do you mean, attracting?

    ONLY if the charges have the same sign. If the charges have opposite signs, then your equation will have an extra negative sign in front of it, and the potential energy of the system will actually decrease when the charges get closer. It's like a ball falling towards the Earth. The potential energy of the system (Earth + ball) decreases as their separation decreases. It's just slightly more complicated with electrodynamics, because there are two different types of charge. A good way to keep track of what will happen is to remember that the system always "wants" to reduce its potential energy, and that will determine how the charges move.
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