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Homework Help: How to visualize a potential difference?

  1. May 19, 2010 #1
    A proton (at rest) is accelerated through a potential difference of 100 v, and it gains speed... how can I picture this in my head so i can understand it?

    My idea:
    Potential difference is the difference between two points a and b. So the proton is traveling from point a (where the potential is 0) to point b (where the potential is 100). But what doesn't add up is... if the proton is going towards a higher potential, then it must be gaining potential energy and losing kinetic energy; but the question above says it gains speed... which means the kinetic energy is increasing.
     
    Last edited: May 19, 2010
  2. jcsd
  3. May 19, 2010 #2
    You have almost got it, let me see if I can help you. Your points A and B, A is where the electron starts, because the potential difference ( energy given to each charge that passes through ) is 100V the electron has potential energy at A ( because it wants to travel to B ). Now as the electron gets closer it is gaining more kinetic energy along the way, speeding up.

    Effectively A is a negative terminal and B is a positive, since electrons are negative they are attracted to the positive terminal and so will move towards it, the attraction will get stronger as they come closer together so the electron speeds up.
     
  4. May 19, 2010 #3
    That actually makes perfect sense... but I thought the potential difference is calculated by V(final) - V(initial). By your definition, the electron starts at A and then goes to B. Therefore, the potential difference = B - A = 0-100 = -100.

    Is -100 potential difference same as +100 potential difference?
     
  5. May 19, 2010 #4
    Electric potential is defined by the electric field. If you have an electric field present, then the electric potential is defined to be V = -[tex]\int E \bullet dl[/tex]. If you have a positive charge, say a proton, then it will feel a force when placed in an electric field and the force will be in the direction of the field.

    Potential energy is defined to be the difference in the potential between two points V(final) - V(initial), just as you said. If a proton moves from a point where the potential is V to a point of lower potential, then it gains energy, as can be seen from the formula I just mentioned. V(final) is smaller than V(initial) there for the potential energy is negative. It is this energy that is converted into the kinetic energy of the proton, from the conservation of energy.

    So going back to your original question. A proton accelerated through a potential difference of 100V, from a point A to a point B, would mean that the potential at A is 100V higher than the potential at B. I think you stated this backwards. If it were to go from A to B where B is 100V higher than A, then it would actually lose energy, in other words it would take energy put in by an external source to move it against the potential gradient to a point of higher potential. So the sign of the potential difference does matter when you're talking about whether a particle gained or lost energy.
     
  6. May 19, 2010 #5
    that makes sense... but if the particle is going from A to B, and gaining kinetic energy along the way, then B has to have a lower potential than A (as you said)... but then the potential difference B-A would be a negative number instead of a positive number..
     
  7. May 19, 2010 #6
    Exactly, think about what you just said. If the potential energy you calculate is negative, that means the particle lost potential energy. The potential energy that was lost had to be gained as kinetic energy. Look at it this way, E= U + K where E is the total energy, U is the potential energy and K is the kinetic energy. From the conservation of energy we know that E is constant, if kinetic energy is lost then potential energy has been gained by the particle and vice versa.
     
  8. May 20, 2010 #7

    Redbelly98

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    They are giving the magnitude of the potential difference. The fact that the proton gained speed tells us that it's a drop of 100V.

    Your understanding is correct.
     
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