Question

the graph shows the electric - field as a function of position in a particular region of space. if $e_x = 600\frac{v}{m}$, what is the potential difference between $x = 0.00\\ m$ and $x = 2.00\\ m$?

Explanation:

Step1: Recall potential - electric field relation

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 region into triangles

We divide the region from $x = 0$ to $x = 2m$ into two right - angled triangles.
The first triangle is from $x = 0$ to $x = 1m$ with base $b_1 = 1m$ and height $h_1=E_0 = 600V/m$. The area of the first triangle $A_1=\frac{1}{2}\times b_1\times h_1=\frac{1}{2}\times1\times600 = 300V$.
The second triangle is from $x = 1m$ to $x = 2m$ with base $b_2 = 1m$ and height $h_2=- 600V/m$. The area of the second triangle $A_2=\frac{1}{2}\times b_2\times h_2=\frac{1}{2}\times1\times(-600)=-300V$.

Step3: Calculate the potential difference

The potential difference $\Delta V=-(A_1 + A_2)$.
$\Delta V=-(300-300)=0V$.

Answer:

$0V$