Answer:
As electrons are negatively charged, they have tendency to go to regions of high potential. As clear from the graph, For \[A\to B\to C\] As \[E=\frac{dV}{dr}=\frac{-({{V}_{A}}-{{V}_{B}})}{d}=\frac{{{V}_{A}}-{{V}_{b}}}{d}\] \[\therefore \] \[{{V}_{A}}-{{V}_{B}}=ED\]
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