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Electric Potential and Fields

  1. Mar 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Which one of the following statements best explains why it is possible to define an electrostatic potential in a region of space that contains an electrostatic field?
    A) Work must be done to bring two positive charges closer together.
    B) Like charges repel one another and unlike charges attract one another.
    C) A positive charge will gain kinetic energy as it approaches a negative charge.
    D) The work required to bring two charges together is independent of the path taken.
    E) A negative charge will gain kinetic energy as it moves away from another negative charge.

    2. Relevant equations


    3. The attempt at a solution

    I Thought it would be 'Like charges repel one another and unlike charges attract one another' because in our text book it states 'In this drawing a positive test charge +q, is situated at Point A between two oppositely charged plates. Because of the charges on the plates, an electric Field E exists in the region between two plates'. But that answer was wrong.

    The correct answer is D but I completely do not understand! Can someone explain it to me please!
     
  2. jcsd
  3. Mar 16, 2014 #2

    vela

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    Why would the fact that the charged plates produce an electric field explain why it's possible to define a potential? When is it possible to define a potential? Try figuring out the answer to the latter question. You might need to go back to an earlier chapter, where you first started learning about the concept of potential energy.
     
  4. Mar 16, 2014 #3
    Well the Formula for Potential Energy is Mass*Gravity*Height...would Height have anything to do with the path taken?
     
  5. Mar 16, 2014 #4

    rude man

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    If I go from the base of a hill to the top of the hill by several different trails, is the height change not the same?
     
  6. Mar 16, 2014 #5

    vela

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    That's the formula for gravitational potential energy. There are other forms of potential energy. What you need to understand is what they have in common. In other words, why can we define a gravitational potential energy? Why can't we define a potential energy for friction? What's the difference between friction and the gravitational force which allows us to define a potential energy in one case and not the other?
     
  7. Mar 16, 2014 #6
    Ahh! That makes sense! Is this what the answer means? This is why it's D?
     
  8. Mar 16, 2014 #7
    I am confused..I am really bad at physics...I don't really know why we can define a gravitational potential energy and why we cant define potential energy for friction.

    I know theres static and kinetic..but thats pretty much it..
     
  9. Mar 16, 2014 #8
    A good way to think about this problem is the fact that

    W = [itex]\intF*dS[/itex]

    This can work out pretty nicely if you think about it in terms of work and energy. The only time the voltage will do work on your particle is when your particle is moving in a motion perpendicular to your field lines. This isn't too hard to visualize conceptually. Think about moving a particle parallel with your field lines. Since your voltage depends on the distance from your other charged object, following a field line would never change anything about your voltage, nor the force felt by that particle. This means that the only relevant quantity to worry about when factoring in your voltage difference is the change in absolute distance from the charged object. So, orbiting a charged sphere with a constant radius R means that your particle will never feel any different force, and the voltage will ALWAYS be constant. As soon as that object starts decreasing that radius, things change.

    So, if you substitute F = q*E into your work equation, you will find that W = qΔV, and since V only depends on your change in distance from your charged object, nothing matters about the path that your particle takes. The only thing that matters is the difference between your initial voltage and your final voltage.
     
  10. Mar 16, 2014 #9

    "So, if you substitute F = q*E into your work equation, you will find that W = qΔV,"

    Wouldnt u get W=q*E*ds?
     
  11. Mar 16, 2014 #10
    The definition of a potential field is where work done on a particle only depends on the initial and final locations in the field and not on the path taken.

    In a gravitational or electric field the force acting on a particle always acts in the same direction. Therefore work done by the field on a particle can be positive or negative depending only on the sign of the direction that the particle moved.

    So if a particle moved 2 positive units in one direction and then back 2 units the total work done by the field would be +2 -2 = 0. If a particle moved 4 units and then 4 units back the work would still be zero.

    Therefore the gravitational and electric fields meet the definition of a potential since the final work done is always the same regardless of how the particle moved.

    A friction force on the other hand always acts against the direction that a particle moves. Therefore the work done by friction never changes sign and is always negative.

    So if a particle moved 2 positive units in one direction and then back 2 units the total work done by friction would be -2 -2 = -4 units. If the particle moved 4 units and then 4 units back the work done would be -8 units.

    Therefore the friction force does not meet the definition of a potential because the work done changes depending on how the particle moved.
     
  12. Mar 16, 2014 #11

    Yes, you would. After you integrate that, however, you will get W = qEΔs. In this case, recall that the only relevant portion of s (arclength) is the change in radius from our charged object. You could replace s with r, in this case, and it is also important to remember that E = -dV/dr, or, after an integration, V = Er. Substitute V = Er into your work equation, qEΔr, and that is how the

    W= qΔV appeared.
     
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