- #1
humo90
- 13
- 0
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.
[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: