GREPhysics.NET: GR8677 #80

GR8677 #80

Problem

GREPhysics.NET Official Solution

Electromagnetism $\Rightarrow$ }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) $Ayz \neq 0$

Alternate Solutions

casseverhart13 2019-08-23 12:20:38	I really appreciate this problem....thanks.... tree service Orlando Reply to this comment
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. Reply to this comment
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. Reply to this comment

Comments

casseverhart13
2019-08-23 12:20:38

I really appreciate this problem....thanks.... tree service Orlando

Reply to this comment

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.

Reply to this comment

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.

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