1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric potential and velocity

  1. Oct 27, 2008 #1
    1. The problem statement, all variables and given/known data
    http://session.masteringphysics.com/problemAsset/1020825/7/jfk.Figure.21.P56.jpg
    A proton's speed as it passes point A is 4.40×10^4 m/s. It follows the trajectory shown in the figure.

    What is the proton's speed at point B?


    2. Relevant equations

    Knowing this would be very helpful

    3. The attempt at a solution

    No attempt yet...If I knew what formulas to use, that would be very helpful.
     
  2. jcsd
  3. Oct 27, 2008 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    By definition, an electric field is accelerating the charge in the direction of decreasing electric potential. Hence, the charge's kinetic energy (and therefore velocity) is increasing.

    But you don't know the electric field and therefore the force. So how do you figure out the change in kinetic energy? Because you know the potential, which is equivalent information. Think about it in terms of conservation of energy. The proton loses electric potential energy in moving from point A to B, which means that it must gain exactly that amount of kinetic energy. And you know, from the diagram, exactly how much potential energy per unit charge is lost in going from A to B. NOW do you understand what calculations need to be carried out?

    Physics is not about memorizing formulae, it's about understanding and applying concepts. We can teach you equations, but if you don't understand physics, you won't know which ones to use in which situations. ;-)
     
  4. Oct 27, 2008 #3
    Sadly, I still do not understand. This entire chapter confuses me.
     
  5. Oct 27, 2008 #4

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Okay. To summarize what I said in my second paragraph:

    potential energy lost = kinetic energy gained.

    potential energy lost is depicted on the diagram. The lines shown are lines of constant electric potential (like contours). The proton is crossing them, going "downhill" as it were. How much potential energy does it lose?
     
  6. Oct 27, 2008 #5
    40v?
     
  7. Oct 27, 2008 #6

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Right, exactly. The potential difference is 30 V - (-10 V) = 40 V.

    Now, electric potential is potential energy PER UNIT CHARGE (1 volt = 1 joule/coulomb).
    So, given *that* potential difference (40 V) between points A and B, how much does the potential energy of this *particular* charge decrease?
     
  8. Oct 27, 2008 #7
    1 v = 1 j/c

    so...

    40 v = 40 j/c

    40 j/c
     
  9. Oct 27, 2008 #8

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Dude, that's how much energy is lost PER UNIT charge. I asked you for the total amount of potential energy lost (in JOULES! Energy is measured in joules). Hint, we have a certain *amount* of charge passing by, and we know how much energy is lost per UNIT of charge.

    Edit: You need to review the concept of electric potential. Make sure you understand why I am saying what I am saying. Electric potential measures how much potential energy there is at a point PER UNIT charge. I.e. a "test" charge of 1 coulomb sitting at that point would have that potential energy.
     
  10. Oct 27, 2008 #9
    Alright, thanks for the help. I shall try again later after reviewing the material.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electric potential and velocity
Loading...