Voltage

• Bringing a 1C test charge from infinity to a point where 1J of energy gets consumed
  • Not really practical so usually we look at change in potential
• $V=\frac{\mathrm{d}W}{\mathrm{d}q}$ in $\frac{\mathrm{J}}{\mathrm{C}}$
• electromotive force

• Measure of potential around one charge - not potential energy. there is no energy yet

Difference in Voltage

• $\mathrm{\Delta }V=V_B-V_B$
• $$$$

\Delta V & =\frac{kq}{r_{B}}-\frac{kq}{r_{A}} \
& =kq\left( \frac{1}{r_{B}}-\frac{1}{r_{A}} \right)
\end

\begin{align}
\Delta V & =kq\left( \frac{1}{r_{B}}-\frac{1}{r_{A}} \right) \
& =9\times 10^{9}(12\times 10^{-6})\left( \frac{1}{0.7}-\frac{1}{0.5} \right) \
& =-61714\mathrm{V}
\end

$$Energylost,checksout(itdoesthisbyitself,noworknecessary)DropinpotentialIf\$-12\mu \mathrm{C}\$,\$\mathrm{\Delta }V=+61714\mathrm{V}\$Energygained,workneeded,checksoutGaininpotential$$