# Chapter 24 - Electric Potential - Problems - Page 710: 14a

$V_A-V_B=-4500V$

#### Work Step by Step

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, $V_A-V_B=4500-9000$ $V_A-V_B=-4500V$

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