- #1
Niles
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[SOLVED] Finding the electric potential of point charge
I want to find the electric potential at a distance r from a point charge. Here's what I'm doing:
We have
[tex]
V = - \int {E \cdot {\rm{d}}{\bf{l}}}
[/tex]
I find that
[tex]
V = - \int_\infty ^r {k_e \frac{q}{{r'^2 }}{\bf{r}} \cdot {\rm{d}}{\bf{l}}} = - \int_\infty ^r {k_e \frac{q}{{r'^2 }}\cos \left( \phi \right){\rm{dl}}} = \int_\infty ^r {k_e \frac{q}{{r'^2 }}{\rm{dl}}} = - k_e \frac{q}{{r }}
[/tex]
I used that I integrate from infinity to r, so dl is pointing opposite to E. I get a negative potential, which is of course wrong for a point charge. Where's my error?
Homework Statement
I want to find the electric potential at a distance r from a point charge. Here's what I'm doing:
We have
[tex]
V = - \int {E \cdot {\rm{d}}{\bf{l}}}
[/tex]
I find that
[tex]
V = - \int_\infty ^r {k_e \frac{q}{{r'^2 }}{\bf{r}} \cdot {\rm{d}}{\bf{l}}} = - \int_\infty ^r {k_e \frac{q}{{r'^2 }}\cos \left( \phi \right){\rm{dl}}} = \int_\infty ^r {k_e \frac{q}{{r'^2 }}{\rm{dl}}} = - k_e \frac{q}{{r }}
[/tex]
I used that I integrate from infinity to r, so dl is pointing opposite to E. I get a negative potential, which is of course wrong for a point charge. Where's my error?