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What is the curl of a electric field?

by back2square1
Tags: curl, electric, field
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back2square1
#1
Feb7-13, 07:45 AM
P: 13
This should be simple but I know I'm going wrong somewhere and I can't figure out where.
The curl of a electric field is zero,
i.e. [itex]\vec { \nabla } \times \vec { E } = 0[/itex]
Because , no set of charge, regardless of their size and position could ever produce a field whose curl is not zero.

But,
Maxwell's 3rd Equation tells us that,
the curl of a electric field is equal to the negative partial time derivative of magnetic field [itex] \vec {B}[/itex].
i.e. [itex]\vec { \nabla } \times \vec { E } = -\frac { \partial }{ \partial t } \vec { B } [/itex]

So is the curl zero or is it not? If we equate those two equations we get that the time derivative of magnetic field is zero. What's wrong? What am I missing?
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jtbell
#2
Feb7-13, 07:54 AM
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P: 11,782
Quote Quote by back2square1 View Post
The curl of a electric field is zero,
i.e. [itex]\vec { \nabla } \times \vec { E } = 0[/itex]
That should read, "the curl of an electrostatic field is zero," that is, the electric field associated with a set of stationary charges has a curl of zero. In this situation, there is no magnetic field, so ##\partial \vec B / \partial t = 0##.
back2square1
#3
Feb7-13, 09:57 AM
P: 13
Oh. Thanks. Got it. Sometimes things as simple as this slip off.


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