Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
Oh I see, that's much simpler than I realized. Thank you! –  Chris Harris May 13 '12 at 0:16
    
@ChrisHarris: Glad to help. –  user26872 May 13 '12 at 0:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.