# Electric Field/Electric Potential (Gradient Notation)

1. Oct 27, 2007

### PFStudent

1. The problem statement, all variables and given/known data

Hey,

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

Last edited: Oct 27, 2007
2. Oct 27, 2007

### Poop-Loops

I wouldn't use it. I would just leave it as:

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

Or

$${\vec{E}} = {-}{\nabla}{V}$$

3. Oct 28, 2007

### cristo

Staff Emeritus
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 $\vec{E}=-\nabla V$

4. Oct 28, 2007

### PFStudent

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

5. Oct 29, 2007

### PFStudent

Hey,

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