## Fundamentals of Physics Extended (10th Edition)

$V_B-V_A=2.46V$
We can find $V_B-V_A$ as: $V_B-V_A=\frac{W}{e}$ We plug in the known values of $W$ and $e$ in the formula above to obtain: $V_B-V_A=\frac{3.94\times 10^{-19}}{1.6\times 10^{-19}}$ $V_B-V_A=2.46V$