# del operator - partial derivatives

I'm taking a class in Electromagnetism, and I'm learning about the relationships between voltage and an electric field from Faraday-Maxwell equations.

The equation I have trouble with is: $$E = -\nabla V$$ where $E$ is the electric field (a vector), $\nabla$ is the gradient operator, and $V$ is the voltage (a scalar).

Given the voltage you may solve for an electric field. My problem lies with the partial derivatives from the gradient operator. If the electric field is equivalent to the gradient of the voltage, then when solving for the electric field, why is it that we integrate the voltage? I figure that this relates to more of my misunderstanding of mathematics, otherwise I would have posted in a different forum. Thanks for the help!

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By "del", do you mean $\nabla$? –  Arturo Magidin May 13 '12 at 0:09
yes I do. More clearly, I mean the partial derivative of the scalar with respect to all directions. –  Chris Harris May 13 '12 at 0:14

Hint: Actually, the electric field is integrated to get the electric potential, $$\begin{eqnarray*} {\bf E} &=& -\nabla V \\ V &=& -\int {\bf E}\cdot d{\bf l}. \end{eqnarray*}$$ Roughly speaking, the inverse of differentiation is integration.