the graph shows the electric field as a function of position in a particular region of space. if (e_x = 1200…

Explanation:

Step1: Recall the relationship between electric - field and potential

The potential difference $\Delta V$ between two points in an electric - field is given by $\Delta V=-\int_{x_1}^{x_2}E_xdx$. Geometrically, this is the negative of the area under the $E_x - x$ curve between $x_1$ and $x_2$.

Step2: Divide the $E_x - x$ graph into geometric shapes

We can divide the graph of $E_x$ versus $x$ from $x = 0$ to $x = 6m$ into three regions: a triangle from $x = 0$ to $x = 2m$, a rectangle from $x = 2m$ to $x = 3m$, and a trapezoid from $x = 3m$ to $x = 6m$.

Step3: Calculate the area of the triangle

The area of a triangle $A_1$ with base $b_1=2m$ and height $h_1 = E_0$ (where $E_0$ is the magnitude of the electric - field at $x = 0$). The formula for the area of a triangle is $A_1=\frac{1}{2}b_1h_1$. Since $E_x$ at $x = 0$ is $- E_0$ and we know that the area under the curve contributes to the potential difference, $A_1=\frac{1}{2}\times2\times E_0=E_0$.

Step4: Calculate the area of the rectangle

The area of a rectangle $A_2$ with length $l = 1m$ and width $w = E_0$. The formula for the area of a rectangle is $A_2=lw$. So $A_2=1\times E_0 = E_0$.

Step5: Calculate the area of the trapezoid

The formula for the area of a trapezoid is $A_3=\frac{1}{2}(b_1 + b_2)h$, where $b_1$ and $b_2$ are the lengths of the parallel sides and $h$ is the height. Here, $b_1=E_0$, $b_2 = 0$, and $h = 3m$. So $A_3=\frac{1}{2}(E_0+0)\times3=\frac{3}{2}E_0$.

Step6: Sum up the areas

The total area $A$ under the $E_x - x$ curve from $x = 0$ to $x = 6m$ is $A=A_1 + A_2+A_3=\frac{1}{2}\times2\times E_0+1\times E_0+\frac{1}{2}(E_0)\times3=\left(1 + 1+\frac{3}{2}\right)E_0=\frac{2 + 2+3}{2}E_0=\frac{7}{2}E_0$. Given $E_0 = 1200V/m$, then $\Delta V=-\frac{7}{2}\times1200=-4200V$.

Answer:

$-4200V$