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Electric Potential

  1. Nov 28, 2011 #1
    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.
     
    Last edited: Nov 28, 2011
  2. jcsd
  3. Nov 30, 2011 #2
    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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