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GR8677 #80
Problem
 GREPhysics.NET Official Solution Alternate Solutions

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
 nakib2010-04-02 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 senatez2006-10-16 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
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.
 neon372010-11-04 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.
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.
 student20082008-10-12 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.
 Prologue2009-11-06 21:03:53 You can't infer that it is electrostatic, the curl is nonzero.

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LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$

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