Electric field from potential and coordinate

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SUMMARY

The electric field can be derived from the electric potential function V(x,y,z) = 3x² + 2y + 5 using the formula E = -∇V. The gradient of the potential function yields the components of the electric field, specifically the x-component as -6x. For the given point s = (5,3,1), the electric field is calculated as E = -∇V = (-6*5, -2, 0) resulting in E = (-30, -2, 0) N/C. This method is essential for understanding the relationship between electric potential and electric fields.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Knowledge of vector calculus, specifically gradient operations
  • Familiarity with partial derivatives
  • Basic physics concepts related to electromagnetism
NEXT STEPS
  • Study vector calculus, focusing on gradient and divergence
  • Learn about electric field calculations from potential functions
  • Explore the implications of electric fields in electromagnetism
  • Review examples of electric potential in different coordinate systems
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone seeking to deepen their understanding of the relationship between electric potential and electric fields.

axgalloway
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Homework Statement



I am given V(x,y,z) = 3x^2 + 2y + 5 and I am given s = (5,3,1), so what is the electric field?

Homework Equations


V= Es


The Attempt at a Solution


I really have no idea. What I tried:

Derivative of V(x,y,z) = 6x+2
so E = 32 N/C?

Really I just don't understand the math necessary. How is it done?
 
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What you really need to do is take the gradient of the potential function:

[tex]E = - \nabla V[/tex]

You'll find the components of the electric field by taking the partial derivative with respect to each variable. Thus the x-component of the field will be -6x.

See: http://hyperphysics.phy-astr.gsu.edu/Hbase/gradi.html"
 
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