Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook