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Verbatim question for GR8677 #80
Electromagnetism}Gauss Law

Recall the differential form of Gauss' Law \nabla \cdot \vec{E}=\rho. For fields that contain no charges, the equation becomes \nabla \cdot \vec{E}=0. Find the choice with 0 divergence.

(A) A(2y-x) \neq 0

(B) A(-x+x)=0 ... so this is it!

(C) A(z) \neq 0

(D) A(yz+xz) \neq 0

(E) Ayz \neq 0




See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
nakib
2010-04-02 09:01:27
Don't go for calculating all the answer choices for \nabla\cdot E = 0.

Eliminate C, D ~> Each term has a non-zero contribution to the div. but both terms are +ve, they can't possible cancel each other out.

Eliminate E ~> Just one term that has a non-zero contribution to the div.

Eliminate A ~> Each term has a non-zero contribution to the div. but one term has a factor of 2 multiplied to it.

(B) is the correct choice.
Alternate Solution - Unverified
senatez
2006-10-16 20:25:28
The electric field must also be a gradient of a function (the potential). Mixed partials of the potential function must be equal.Alternate Solution - Unverified
Comments
nakib
2010-04-02 09:01:27
Don't go for calculating all the answer choices for \nabla\cdot E = 0.

Eliminate C, D ~> Each term has a non-zero contribution to the div. but both terms are +ve, they can't possible cancel each other out.

Eliminate E ~> Just one term that has a non-zero contribution to the div.

Eliminate A ~> Each term has a non-zero contribution to the div. but one term has a factor of 2 multiplied to it.

(B) is the correct choice.
neon37
2010-11-04 04:40:24
Its not too hard if you think about how to do it fast before hand. It might take longer to think about some other logic. Divergence is a dot product, so for \hat{e_i} terms you take \frac{\partial}{\partial x_i}. That shouldnt be too difficult. In this case it was just the second one, so it took me about 7 seconds maximum after reading the question.
Alternate Solution - Unverified
senatez
2006-10-16 20:25:28
The electric field must also be a gradient of a function (the potential). Mixed partials of the potential function must be equal.
student2008
2008-10-12 12:08:08
Not necessarily. Since \vec E = -\frac1c \frac{\partial\vec A}{\partial t} -\nabla \phi, you're right only in the electrostatic case, which takes place here. And in general we can't get rid of the \frac{\partial \vec A}{\partial t} by a gauge transformation, since the condition \frac{\partial \vec A}{\partial t}=0 implies {\bf three} scalar ones.

Although, since there're only {\bf static} expressions for the electric field, we can infer that the problem {\bf is} electrostatic.
Prologue
2009-11-06 21:03:53
You can't infer that it is electrostatic, the curl is nonzero.
Alternate Solution - Unverified

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