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Is e field equal to the negative derivative of electric potential?

  • Thread starter pyroknife
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  • #1
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Since V is the integral of -E. Shouldn't E be derivative of -V. I asked my TA that and he said something about them not having that kind of a relationship, can someone explain this?
 

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  • #2
kuruman
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You need to sharpen your thinking. The electric field is a vector, so to specify it you need to specify its three components. These are

[itex]E_{x}=-\frac{\partial V}{\partial x}[/itex]; [itex]E_{y}=-\frac{\partial V}{\partial y}; [/itex][itex]E_{z}=-\frac{\partial V}{\partial z}[/itex]
 
  • #3
613
3
You need to sharpen your thinking. The electric field is a vector, so to specify it you need to specify its three components. These are

[itex]E_{x}=-\frac{\partial V}{\partial x}[/itex]; [itex]E_{y}=-\frac{\partial V}{\partial y}; [/itex][itex]E_{z}=-\frac{\partial V}{\partial z}[/itex]
Doesn't that just mean E is the negative derivative of V tho?
 
  • #4
kuruman
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It means that the x component of E is the negative derivative of V with respect to x
and that the y component of E is the negative derivative of V with respect to y
and that the z component of E is the negative derivative of V with respect to z.
 
  • #5
613
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ah okay thanks
 

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