Determining an electric field in vector form, got the answer, just checking.

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SUMMARY

The discussion focuses on calculating the electric flux through a surface using vector notation. The surface area vector is defined as A = (2i + 3j) m², and two uniform electric fields are analyzed: E = 4i N/C and E = 4k N/C. The calculated flux for the first field is 8 Nm²/C, derived from the dot product of the electric field and area vector, while the flux for the second field is 0 due to the absence of a k component in the area vector. The key takeaway is the importance of vector components in determining electric flux.

PREREQUISITES
  • Understanding of vector mathematics and dot products
  • Familiarity with electric field concepts and units (N/C)
  • Knowledge of electric flux and its calculation
  • Basic grasp of surface area vectors in physics
NEXT STEPS
  • Study the concept of electric flux in more detail
  • Learn about vector operations, specifically dot products
  • Explore different configurations of electric fields and their effects on flux
  • Investigate applications of electric fields in real-world scenarios
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Students studying electromagnetism, physics educators, and anyone looking to deepen their understanding of electric fields and vector mathematics.

mr_coffee
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Hello everyone, I finally finished the chapter and now I'm going to attempt to do some problems. Well the first problem says: A surface has the are vector A = (2[itex]\hat i + 3\hat j[/itex]m^2. What is the flux of a uniform electric field through it if the field is (a) E = 4[itex]\hat iN/C[/itex] (b) [itex]E = 4\hat k[/itex]. The answer for (a) is 8 Nm^2/C (b) 0. The answer is 8 for the first one because well (4)(2)...the j component isn't used at all because why? Also b is 0 because k isn't in the same vector as i or j, is that right or am i getting lucky?
 
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There's no j component for a. because the j component of the field you're given is 0. You're taking a dot product to find the flux, ie

[tex]\Phi_{E} = \vec{E} \cdot \vec{A}[/tex]

So, when you evauluate that and you multiply the j components, you get 3 x 0 which is of course 0. That's the same reason that b. is 0, because [itex]\vec{A}[/itex] has no k component and [itex]\vec{E}[/itex] has no i or j components.
 
ohh! i forgot all about that, i have to refresh on vector math! thanks!
 

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