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Electric Potential Energy concepts

  1. Sep 23, 2014 #1
    1. The problem statement, all variables and given/known data

    1. What does the following equation mean and how is it analogous to the relationship between the electric field and the the electrostatic force?

      V ( ~r ) = U ( ~r ) / q
    In general, confused about concepts, meanings and vocabulary. The naming of different terms in this section is annoying to sort out. Voltage, V, is defined as electric potential energy, but is also called electric potential, which also refers to U, which is actually the electric potential. Is this correct?

    Questions I'm trying to understand:
    - How are V and U different?

    - Related to U: Negative charges go from low to high potential in positive E field. How does this work intuitively? They go against the direction of the E field and this brings them closer to the positive charge. Is it because if you keep them apart when they are close it takes a lot of force to do so? (just thought of this, this makes a lot of sense)

    It makes sense for two positive or two negative charges, since you have to hold them there or they will want to separate like a compressed spring.

    - The difference of Vf and Vi is the integral of E dot dr, dr being tangent to the E field at each point. Work is done by the electric force to move particles and this is equal but opposite to the change in electric potential. Is voltage dangerous because if let go (or circuit is connected) a lot of charge will be released?

    - When V is defined from bringing a particle from infinity (i.e. infinite r makes the potential zero), how does what the particle is being brought to effect anything?

    2. Relevant equations

    V = U / q = kQ/r

    U = kQq / r


    3. The attempt at a solution

    What I'm thinking so far: electric potential created by taking charge q close to another charge Q at distance r. Dividing this by q is V, the electric potential energy, which is the work per charge to bring it to that point. The electric field is what you have to work against because of the electrostatic force it exerts on charged particles.

    Thanks for taking the time to read through my confusion if you do
     
  2. jcsd
  3. Sep 23, 2014 #2
    I remember my confusion over u and v back in college! So don't feel bad, you aren't the only one. :)
    I will start with E (electric field) then move to V electric potential and U electric potential energy.
     
  4. Sep 24, 2014 #3
    As you know, according to Coulomb's law, a charge q, that is a distance r away from another charge Q, exerts the force
    [tex]F = k \frac{q Q}{r^2}\ [/tex], Now we want to know what is the force that q exerts on a unit charge at distance r. To find it we simply need to divide F by Q and call the results E. Meaning, [tex]E = \frac{F}{Q}[/tex]. So technically Electric Field (E) is just the force on unit charge (we'll leave it here for now)

    Now about electric potential, unlike electric field which is a vector quantity, electric potential (V) is a scalar, which means it can be easier to work with. Now what is electric potential? It is the amount of electric potential energy a unit charge will have if we put it at a distance r from a charge q. (you can follow electric field analogy here: E is F per unit charge (E=F/q), V is U per unit charge, so V=U/q)
    another way to look at V is to imagine it as the work done in carrying a unit charge from infinity to the point a distance r from a charge q.

    So remember that Electric potential energy (U) and electric potential (V) are not the same things. U would be the work done in carrying a charge Q from infinity to the point a distance r from a charge q, and V is the work done in carrying a unit charge from infinity to the point a distance r from a charge q (or the work per unit charge)
    We usually need to know only the difference between electric potentials of two point, which can be calculated using the following equation:
    [tex] \Delta V = V_{b} - V_{a} = - \int_a^b\vec{E}\cdot d\vec{l} [/tex]

    Now about the direction negative charges move: Imagine we have a uniform electric field in the direction of x, (from left to right). What will be the force on a negative charge in this field? as we said before F=Q.E since Q in negative, the force F would be in the opposite direction of E, (hence from right to left) which means the charges will be moved from right to left in the direction of -x.
    we can calculate deltaV using the above equation. Here E is in x direction, and dl is in -x direction (we just proved why), which means the angle between them is 180, so E.dl will be negative, and considering the minus sign in deltaV equation, the difference between electric potentials will be a positive number, which means the negative charges are moving from lower to higher potential.
     
  5. Sep 24, 2014 #4
  6. Sep 24, 2014 #5
    Thank you for the response. I'm good with the directions and dot product, that makes sense. The difference between U and V is similar between F and E, so does this mean V is the potential of the field measured by a unit charge that doesn't effect the field? With U being the energy put into moving the specific charge Q

    Also the MIT lectures are very good.
     
  7. Sep 24, 2014 #6
    yes, that's right! :)
     
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