Graphing the Curl of -Cx\vec{z} on the xz plane

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The discussion focuses on finding the curl of the vector field E = -Cx\hat{z}. The curl is calculated as ∇xE = C\hat{y}, indicating that it is constant throughout the xz plane. The original poster questions whether they missed any components in their derivatives, but it is confirmed that the curl remains constant and does not vary across the plane. The key takeaway is that the curl of -Cx\hat{z} results in a uniform vector field in the y-direction. This understanding is crucial for accurately graphing the curl on the xz plane.
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I posted the divergence of this earlier but thought I should post the curl separately.

Homework Statement



Find the curl of ##E=-Cx\hat{z}##


Homework Equations


∇xE=##[\frac{∂E_z}{∂y}-\frac{∂E_y}{∂z}]\hat{x}+[\frac{∂E_x}{∂z}-\frac{∂E_z}{∂x}]\hat{y}+[\frac{∂E_y}{∂x}-\frac{∂E_x}{∂y}]\hat{z}##


The Attempt at a Solution



Since there's only a z component
∇xE=##-\frac{∂E_z}{∂x}=C\hat{y}##

I'm suppose to graph this onto the xz plane. But, isn't all the same throughout the plane? I feel like maybe I missed a component from the derivatives but I think all the rest are 0, right?

Thanks for any guidance.
 
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Yes, that is correct. For ever point the curl of -Cx\vec{z} is the constant C\vec{y}.
 
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