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

[pyroknife]

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?

 

[kuruman]

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

$E_{x}=-\frac{\partial V}{\partial x}$; $E_{y}=-\frac{\partial V}{\partial y};$$E_{z}=-\frac{\partial V}{\partial z}$

 

[pyroknife]

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

$E_{x}=-\frac{\partial V}{\partial x}$; $E_{y}=-\frac{\partial V}{\partial y};$$E_{z}=-\frac{\partial V}{\partial z}$

Doesn't that just mean E is the negative derivative of V tho?

 

[kuruman]

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.

 

[pyroknife]

ah okay thanks