Answer
1.8J
Work Step by Step
$V_A=k(\frac{38\mu C}{0.12m}+\frac{38\mu C}{0.24m})=4.75\times10^6J$
$V_B=k(\frac{38\mu C}{0.184m}+\frac{38\mu C}{0.278m})=3.43\times10^6J$
$W=qV_{ba}=(-1.5\mu C)(3.43\times10^6J-4.75\times10^6J)=1.8J$
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