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

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SUMMARY

The electric field (E) is defined as the negative gradient of the electric potential (V), represented mathematically as E_{x}=-\frac{\partial V}{\partial x}, E_{y}=-\frac{\partial V}{\partial y}, and E_{z}=-\frac{\partial V}{\partial z}. This indicates that each component of the electric field corresponds to the negative derivative of the electric potential with respect to its respective spatial coordinate. The relationship is indeed valid, confirming that E is the negative derivative of V in a vector form.

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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?
 
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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}
 
kuruman said:
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?
 
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.
 
ah okay thanks
 

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