Question

the graph shows the electric field as a function of position in a particular region of space. if (e_x = 400\frac{v}{m}), what is the potential difference between (x = 2.00 m) and (x = 4.00 m?)

Explanation:

Step1: Recall the relationship between electric - field and potential

The relationship between the electric - field $E_x$ and the potential difference $\Delta V$ is given by $\Delta V=-\int_{x_1}^{x_2}E_xdx$. In a region where the electric - field is constant, $\Delta V=-E_x\Delta x$.

Step2: Identify the electric - field value in the relevant region

From the graph, for the region between $x = 2.00m$ and $x = 4.00m$, the electric - field $E_x=- 400\frac{V}{m}$. The negative sign indicates the direction of the electric - field.

Step3: Calculate the potential difference

We know that $\Delta x=x_2 - x_1$, where $x_1 = 2.00m$ and $x_2 = 4.00m$, so $\Delta x=4.00 - 2.00=2.00m$. Using the formula $\Delta V=-E_x\Delta x$, substituting $E_x=-400\frac{V}{m}$ and $\Delta x = 2.00m$, we get $\Delta V=-(-400\frac{V}{m})\times2.00m$.
$\Delta V = 800V$.

Answer:

$800$