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Gradient equal to multiplying by vector?

  1. Oct 10, 2015 #1
    Hi guys!
    So I have been researching the electric field, and I have come upon some interesting equations that confused me a little (all from wikipedia): fe40e6e5030963130cd36c93ca29f634.png , 400073dd9ef3633c03d01c35c9f6281e.png , and 8757428f7aeb787795811caa40897baa.png with V being the same as psi. Doing the algebra, I would get (Q/4πε0)*(r-hat/|r|2)=-∇(Q/4πε0r)-∂A/∂t. Now in the case that A does not change with time, we would get that taking the negative gradient is equal to multiplying by r-hat/|r|. Is this correct?
  2. jcsd
  3. Oct 10, 2015 #2
    Yes, this is true for the gradient of ##\frac{1}{|r|}##. $$-\nabla \frac{1}{|r|} = \frac{\hat{\mathbf{r}}}{|r|^2}, \;\;\;\;\;\;\;\;r\neq0.$$
  4. Oct 11, 2015 #3
    Would the negative gradient of 2|r|\frac{2}{|r|} be 2r-hat over |r|2?
  5. Oct 11, 2015 #4
    Sorry, I don't know what happened with the code. I was asking bout 2/|r|
  6. Oct 11, 2015 #5
    you can always take a moltiplicative constant out of the gradient.

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