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Gradient of a potential energy function

  1. Jan 26, 2014 #1

    Radarithm

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    1. The problem statement, all variables and given/known data
    Find the derivative of [tex]\frac{Q}{4\pi \epsilon_0 r}[/tex]


    2. Relevant equations
    [tex]\frac{d}{dx} \frac{1}{x}=\ln x[/tex]


    3. The attempt at a solution

    Assuming [itex]Q[/itex] and the rest of the variables under it are constant, [tex]\frac{Q}{4\pi \epsilon_0}\frac{1}{r}[/tex] then the derivative should be [itex]\ln r[/itex]. I am taking the gradient of a potential energy function but since it is in one dimension ([itex]r[/itex] in this case isn't a 2-3 dimensional vector) it is the same as taking the derivative. Is my answer correct or did I make a mistake somewhere?
     
  2. jcsd
  3. Jan 26, 2014 #2

    Mentallic

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    You have it the wrong way around.

    [tex]\frac{d}{dx}\ln{x}=\frac{1}{x}[/tex]
     
  4. Jan 26, 2014 #3

    Radarithm

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    Yep, thanks for correcting me. The derivative is then [tex]\frac{-1}{r^2}[/tex] from the power rule, correct?
     
  5. Jan 26, 2014 #4

    Mentallic

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    Correct.
     
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