# Gradient equal to multiplying by vector?

1. Oct 10, 2015

### Isaac0427

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): , , and 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. Oct 10, 2015

### MisterX

Yes, this is true for the gradient of $\frac{1}{|r|}$. $$-\nabla \frac{1}{|r|} = \frac{\hat{\mathbf{r}}}{|r|^2}, \;\;\;\;\;\;\;\;r\neq0.$$

3. Oct 11, 2015

### Isaac0427

Would the negative gradient of 2|r|\frac{2}{|r|} be 2r-hat over |r|2?

4. Oct 11, 2015

### Isaac0427

Sorry, I don't know what happened with the code. I was asking bout 2/|r|

5. Oct 11, 2015

### lightarrow

you can always take a moltiplicative constant out of the gradient.

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lightarrow