Answer
$(V_B-V_A)=-157V$
Work Step by Step
According to the given scenario:
$W_{electric}+W_{external}=KE_{final}-KE_{initial}$
$-q(V_B-V_A)+W_{external}=KE_{final}-0$
This simplifies to:
$(V_B-V_A)=\frac{W_{external}-KE_{final}}{q}$
We plug in the known values to obtain:
$(V_B-V_A)=\frac{15.0\times 10^{-4}-4.82\times 10^{-4}}{-6.50\times 10^{-6}}\approx-157V$
