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Electrostatics and potential question

  1. Jan 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A 2.5C charge is moved from a point with a potential of 12V to another point of potential 75V. How much work was done on this charge?


    2. Relevant equations
    ΔV = ΔEp/Q


    3. The attempt at a solution
    I am unsure how to do this question but what I did do is I took the two potentials and I subtracted the two to get a difference in potential. Then I took that difference in potential and divided it by the charge of 2.5C to get the change in voltage.

    Now that I have the change in voltage what equation am I supposed to use to find the amount of work done?
     
  2. jcsd
  3. Jan 3, 2009 #2

    jambaugh

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    The change in electric potential is the change in voltage. Your equation would better be written [itex] Q\Delta V = \Delta E[/itex]. What is the p in your equation?
     
  4. Jan 3, 2009 #3
    Are you talking about the "p" in ΔEp? If so then the "p" together with the "E" stands for electric potential (because that's how it is written in the formula sheet that is given to us).

    So I take the two point potentials and find the difference between them and multiply that by the charge to get ΔEp (I tried that and got a ΔEp= 63)? Then what would I do with that ΔEp?
     
  5. Jan 4, 2009 #4

    Defennder

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    What do you understand by 'ΔEp' in your given equation?
     
  6. Jan 4, 2009 #5

    jambaugh

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    I don't know who wrote your formula sheet but it appears to be wrong. If you define [itex]\Delta E[/itex] as the change in energy and [itex]\Delta V[/itex] as the difference in voltage (electric potential) and drop the p then your formula would be right.

    Look at the units, look at the units, look at the units!

    1 Volt = 1 Joule per Coulomb. You have the formula built into the units.
    Volts times Coulombs equals Joules
    Voltage difference times charge equals energy!

    w.r.t. 63 check your arithmetic and keep your units. The difference in potentials is 75volts - 12volts = 63volts. Multiply that by the charge to get the energy.

    I'll say it some more... units units units units units units.... and I can't emphasize enough UNITS! Raw numbers are meaningless. If you want to understand the physics you MUST pay attention to units. If you want to do well on tests you really need to understand the physics.

    Sorry to vent at you but this is one of my biggest gripes with my students. They drop units and loose any track of what they're doing. Remember that you aren't saving time or work if you have to repeat the attempts over and over. Work it once with the units carefully included and you'll get there more quickly and with, what is more important, confidence in your answer.
     
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