Answer
(a) region 4, region 4
(b) $1: 13V/m; 2: 0; 3: -5.1V/m; 4: 68V/m $
Work Step by Step
(a) We know that $ E=-=\frac{\Delta V}{\Delta S}$
$\Delta V $ is a change in electric potential and $\Delta S $ is the displacement. The electric field is opposite of the slope. Thus, the value of $ E $ is maximum when the slope has maximum negative value. We conclude that in region 4 the value of $ E_x $ and its magnitude will be at a maximum.
(b) In region 1, the electric field along the x direction is$ E_{x1}=-\frac{\Delta V}{\Delta S}$
$ E_{x1}=-\frac{-(6V-8V)}{0.15m-0m}=13V/m $
In region 2
$ E_{x2}=\frac{-\Delta V}{\Delta S}$
$ E_{x2}=\frac{-(6V-6V)}{0.25m-0.15m}=0$
In region 3, $ E_{x3}=\frac{-\Delta V}{\Delta S}$
We plug in the known values to obtain:
$ E_{x3}=\frac{-(7.8-6V)}{0.6m-0.25m}=-5.1V/m $
In region 4, $ E_{x4}=\frac{-\Delta V}{\Delta S}$
We plug in the known values to obtain:
$ E_{x4}=\frac{-(1V-7.8V)}{0.7m-0.6m}$
$ E_{x4}=68V/m $
