Electric Potential: Dealing with Vectors in Integral Boundaries

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SUMMARY

The discussion centers on calculating electric potential using vector notation in integral boundaries, specifically the expression V(r) = -∫E·ds. The electric field E is defined as E = (ρR³)/(3ε₀z²)âˆhat{z}, and the differential path element is ds = dzâˆhat{z}. The confusion arises when substituting the upper limit b as a vector instead of a scalar, leading to the conclusion that the potential should be expressed as V(z) rather than V(r), as only the z-component is relevant in this context.

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  • Understanding of electric potential and electric fields
  • Familiarity with vector calculus and integral boundaries
  • Knowledge of the concepts of scalar and vector quantities in physics
  • Basic proficiency in evaluating definite integrals
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This discussion is beneficial for physics students, electrical engineers, and anyone studying electromagnetism, particularly those dealing with electric potential and vector calculus.

humo90
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I am confusing about dealing with the vectors in integral boundaries of the electric potential;
[itex]^{b}_{a}[/itex]∫E.ds where a and b are vectors.
For example, if I would calculate the potential for outside region of a sphere along z-direction, I would use E=[itex]\frac{ρR^3}{3ε_{0}z^2}[/itex][itex]\hat{z}[/itex], and ds=dz[itex]\hat{z}[/itex]
then V(r)=-[itex]^{b}_{∞}[/itex]∫[itex]\frac{ρR^3}{3ε_{0}z^2}[/itex][itex]\hat{z}[/itex].dz[itex]\hat{z}[/itex] = -[itex]^{b}_{∞}[/itex]∫[itex]\frac{ρR^3}{3ε_{0}z^2}[/itex].dz
After evaluating the integral which would be V(r)=[[itex]\frac{ρR^3}{3ε_{0}z}[/itex]][itex]^{b}_{∞}[/itex], say b=b[itex]\hat{z}[/itex], if I plug in b as magnitude the result would be as usual, but if b is vector, then how could I plug it in this potential function? Please help.
 
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Why do you plug the vector in it when you have just one component z. It is not V(r) but V(z), so that the vector is not needed.
 

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