Answer
a) 0 V
b) -48 V
c) 60 V
d) No
Work Step by Step
(a) We know that
$\Delta V=E(\Delta x)$
We plug in the known values to obtain:
$\Delta V=(1200N/C)(0.0m)=0V$
(b) As $\Delta V=-E(\Delta x)$
We plug in the known values to obtain:
$\Delta V=-(1200N/C)(0.04m)=-48V$
(c) We know that
$\Delta V=E(\Delta x)$
We plug in the known values to obtain:
$\Delta V=(1200N/C)(0.05m)=60V$
(d) We can see from the above results that we only have a potential difference at any two points and it is not possible to find the potential ($V_A$) at point A (it can be determined only when we are given either $V_B$ or $V_C$).