# Electric Field/Electric Potential (Gradient Notation)

 P: 171 1. The problem statement, all variables and given/known data Hey, I have a question about Electric Field/Electric Potential gradient notation. Since, $${\vec{E}} = {-}{\nabla}{V(r)}$$ Which reduces to, $$\vec{E} = {-}{\nabla}{V(x, y, z)}$$ When expanded is, $$\vec{E} = {-}{\left[{\frac{\partial[V]}{\partial{x}}}{\hat{i}} + {\frac{\partial[V]}{\partial{y}}}{\hat{j}} + {\frac{\partial[V]}{\partial{z}}}{\hat{k}}\right]}$$ So using partial derivative notation can I write, $${\vec{E}} = {-}{\vec{V}'_{xyz}}$$ So, is the above correct notation? The reason I am hesitant is, because formally the gradient is defined as a vector operator that takes a scalar field (such as the electric potential) and changes it to a vector field (such as the electric field) through: partial differentiation with the addition of unit vectors ($$\hat{i}, \hat{j}, \hat{k}$$). However, writing it as below sort of implies the potential is a vector (which it isn't), but gives the impression that it is because of how the gradient is defined. $${\vec{E}} = {-}{\vec{V}'_{xyz}}$$ So, is the above notation correct? -PFStudent
P: 863
 Quote by PFStudent However, writing it as below sort of implies the potential is a vector (which it isn't), but gives the impression that it is because of how the gradient is defined. $${\vec{E}} = {-}{\vec{V}'_{xyz}}$$ So, is the above notation correct? -PFStudent
I wouldn't use it. I would just leave it as:

$${\vec{E}} = {-}{\nabla}{V(r)}$$

Or

$${\vec{E}} = {-}{\nabla}{V}$$
Mentor
P: 8,317
 Quote by PFStudent $${\vec{E}} = {-}{\vec{V}'_{xyz}}$$ So, is the above notation correct?
No, it is not correct. That is, there is no notation I know of that looks like that that is defined as the gradient of a scalar field.

As the above poster says, there is nothing wrong with $\vec{E}=-\nabla V$

 P: 171 Electric Field/Electric Potential (Gradient Notation) Hey, Yea, thanks for the input, I can see why that notation, $${\vec{E}} = {-}{\vec{V}'{xyz}}$$ is wrong. Since, we are adding the components of a vector that is not the same as taking the partial derivative of a function with respect to each of the variables. Since, all the gradient is doing is the following, $$\vec{E} = {-}{\nabla}{V(x, y, z)} = {-}{\left[{\frac{\partial}{\partial{x}}{\left[V\right]}} + {\frac{\partial}{\partial{y}}}{\left[V\right]}} + {\frac{\partial}{\partial{z}}}{\left[V\right]}}\right]}{\hat{r}}$$ Thanks, -PFStudent
 P: 171 Hey, I've been thinking about this and I have a follow up question. Since, $$\vec{E} = {-}{\nabla}{V(r)} = {-}{\left[{\frac{\partial}{\partial{x}}{\left[V\right]}} + {\frac{\partial}{\partial{y}}}{\left[V\right]}} + {\frac{\partial}{\partial{z}}}{\left[V\right]}}\right]}{\hat{r}}$$ and also, $$E = {-}{\frac{\partial}{\partial{r}}}{\left[{V(r)}\right]}$$ So then, $$\vec{E} = {-}{\nabla}{V(r)} = {-}{\frac{\partial}{\partial{r}}}{\left[{V(r)}\right]}{\hat{r}}$$ Now can I rewrite the above as below? $$\vec{E} = {-}{\frac{\partial}{\partial{(x, y, z)}}}{\left[{V(x, y, z)}\right]}{\hat{r}}$$ Which for $${E}$$ can also be written as, $${E} = {-}{\frac{\partial}{\partial{(x, y, z)}}}{\left[{V(x, y, z)}\right]}$$ So, is the notation for the above two equations correct? Thanks, -PFStudent

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