- #1

tshafer

- 42

- 0

[tex]

\vec E = \frac{Q}{4\pi\varepsilon_0} \frac{\hat r}{r^2}.

[/tex]

Finding the potential should be trivial... I do the integral

[tex]

V = -\int_a^b \vec E \cdot d\vec r = -\frac{Q}{4\pi\varepsilon_0} \int_a^b \frac{dr}{r^2} = -\frac{Q}{4\pi\varepsilon_0}\left( \frac{1}{a} - \frac{1}{b} \right).

[/tex]

This is off by a minus sign and I cannot understand why... reversing the limits of integrations, of course, kills the sign but introduces a sign flip in the dot product as E then opposes dr. This should be trivial... help? :)