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Homework Help: Electric Field vs. Electric Potential

  1. Feb 26, 2010 #1

    JJBladester

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    1. The problem statement, all variables and given/known data

    What is the difference between an electric field and electric potential?

    2. Relevant equations

    Electric Field = E = Kq/r2

    Electric Potential = V = Kq/r

    3. The attempt at a solution

    From the equations, I can see that the electric field strength decreases as an inverse square of r (distance) whereas the electric potential decreases linearly with distance.

    The electric field is a vector while the electric potential is a scalar.

    An electric field is the sum of any electric forces acting on a charge or group of charges. An electric field can cause a particle (proton, electron) to accelerate, as when passing through a charged capacitor in a CRT display. Flux is the amount of, or flow of electric field passing through an object.

    Both electric field and electric potential can be obtained by superposition (adding up individual field contributions or individual potential contributions).

    All this is great and taken from various pages in my book... But, what does it all mean? I still don't have a clear *intuition* about the difference between electric field and electric potential.

    I know a battery has electric potential (voltage) and there is an electric field within it between the positive and negative sides. What else can you add to help me realize the *physical* concept of electric field vs. electric potential?
     
  2. jcsd
  3. Feb 26, 2010 #2
    the voltage at a point represents the work per unit charge needed to place any charge q there. it is analogous to saying gravitational potential is the work per unit mass needed to place mass at a height h. (so gravitational potential would be g*h and you multiply any mass into it to find gravitational potential energy. With electric potential, you multiply q into E*d to find the electric potential energy. Notice that E and G are both fields and that h and d are both distances)

    E field is a vector that represents how much force and what direction that force will be per unit charge placed in that field. This is analogous to saying g is the force per unit mass for any mass at a point inside a gravitational field of g. Remember this from mechanics: FORCE = ma. Well, force also = qE
     
  4. Feb 26, 2010 #3

    collinsmark

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    Electric potential is, in a sense, a measure potential energy per unit charge. The electric field is a measure of force per unit charge.

    Electrical potential, like anything dealing with potential energy, there is assumed to be some arbitrary/chosen reference point. For example, you could determine the potential energy of a cart on a roller-coaster. But you'd have to pick a relative reference point for comparison. The ground would be a common pick (making all your potential energies relative to the ground, where the coaster stands). But you could also pick the center of the earth as your reference point, or 1000 feet above if you wanted to. Whatever choice you make it doesn't change the relative potential energies of different heights of the roller-coaster cart, but it is always assumed that there is such a reference point. When specifying the voltage of a battery, the assumption is one terminal in reference to the other terminal. Often, electric potentials are referenced with respect to infinity. But that's merely a useful convention.

    The electric field, which involves force, doesn't necessarily need such a reference point. Forces are capable of standing on their own.

    You mentioned that electric fields are vectors, and electric potentials are scalars. That makes a big difference (appreciate this). I'll try to explain why. Examine both the equations for the electric field and electric potential (with respect to infinity) of a point charge, in a rigorously more detailed form:

    [tex] \vec E = \frac{1}{4 \pi \epsilon_0}\frac{q}{r^2}\hat r [/tex]

    [tex] V = \frac{1}{4 \pi \epsilon_0}\frac{q}{r} [/tex]

    Yes, there's the difference of the 1/r2 vs 1/r that you pointed out. But perhaps more importantly is that pesky [tex] \hat r [/tex] unit vector in the electric field equation. This makes using superposition to add up multiple electric fields a real pain. For every point in the electric field, you must add up all the respective components separately. This is especially painful, because the constituent electric fields from different charges generally each have different unit vectors of their own.

    For the above reason, it's often easier to calculate the electric fields of a complicated system by first converting everything to electric potentials as an interim step. Then use superposition to add the potentials together in happy scalar land. Lastly go back to the electric field using the gradient of the sum.

    Do you remember in earlier in your studies where you were dealing with kinematics, and all these complicated equations dealing with force, acceleration, velocity and position. Everything was vector based, and sometimes it got complicated, going back and forth with all these equations and vectors? But then you found you could figure out the same think using conservation of energy, making the problem a cinch? Well electric potentials aren't exactly the same thing, but they are a big step in that direction.
     
  5. Mar 1, 2010 #4

    JJBladester

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    Thank you both. Your explanations have made a big impact on my intuitive grasp of the two topics mentioned above. Have a great week!
     
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