- #1
humo90
- 13
- 0
I am confusing about dealing with the vectors in integral boundaries of the electric potential;
^{b}_{a}∫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=\frac{ρR^3}{3ε_{0}z^2}\hat{z}, and ds=dz\hat{z}
then V(r)=-^{b}_{∞}∫\frac{ρR^3}{3ε_{0}z^2}\hat{z}.dz\hat{z} = -^{b}_{∞}∫\frac{ρR^3}{3ε_{0}z^2}.dz
After evaluating the integral which would be V(r)=[\frac{ρR^3}{3ε_{0}z}]^{b}_{∞}, say b=b\hat{z}, 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.
^{b}_{a}∫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=\frac{ρR^3}{3ε_{0}z^2}\hat{z}, and ds=dz\hat{z}
then V(r)=-^{b}_{∞}∫\frac{ρR^3}{3ε_{0}z^2}\hat{z}.dz\hat{z} = -^{b}_{∞}∫\frac{ρR^3}{3ε_{0}z^2}.dz
After evaluating the integral which would be V(r)=[\frac{ρR^3}{3ε_{0}z}]^{b}_{∞}, say b=b\hat{z}, 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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