## Fundamentals of Physics Extended (10th Edition)

$V_A-V_B=-4500V$
We know that $V_A=K\frac{q}{d_1}$. First, we find $V_A$: $V_A=9\times 10^9\times \frac{1\times 10^{-6}}{2}=4500V$ Next, we find $V_B$: $V_B=K\frac{q}{d_2}$ $V_B=9\times 10^9\times \frac{1\times 10^{-6}}{1}=9000V$ Now, $V_A-V_B=4500-9000$ $V_A-V_B=-4500V$