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## GR0877 Problem 91

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (C) When $\nabla \times \mathbf{E} = 0$, the line integral of $\mathbf{E}$ around any closed loop, according to Stokes’ theorem, is zero. Because of this, we can unambiguously talk about a scalar function $\displaystyle \varphi(r) = - \int_{\mathcal{O}}^{\mathbf{r}} \mathbf{E} \cdot d\mathbf{l}$ such that $\mathbf{E} = - \nabla \varphi$.

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1. 29.02.2012 at 01:53

We know that the curl of a gradient is always 0. Given E is the gradient of a scalar, the curl of E must be 0!

• 16.10.2013 at 21:30

Be careful. The curl of E isn’t always zero. Recall your Maxwell Equations where
$latex \nabla \times E=-\frac { \partial B }{ \partial t }$
which of course can be non-zero if there’s a time-varying magnetic field. But I guess it must be to have a unique solution to $latex E=- \nabla \phi$ .

• 16.10.2013 at 21:31

damn! my latex didn’t work! admin, can you please fix it? and thank you, generally, for the wonderful site!

2. 16.10.2013 at 21:34

Hm… Trying again:
Be careful. The curl of E isn’t always zero. Recall your Maxwell Equations where
$\nabla \times E=-\frac { \partial B }{ \partial t }$
which of course can be non-zero if there’s a time-varying magnetic field. But I guess it must be to have a unique solution to $E=- \nabla \phi$ .