Question

Figure EX26.3 is a graph of $E_x$. What is the potential difference between $x_i=1.0m$and $x_f=3.0m$?

Short answer

The required potential difference is$\mathbf{}\mathbf{∆}\mathit{V}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}\mathit{V}$

Step-by-step solution

Given information and formula used

Given :

$x_i=1.0m$ and $x_f=3.0m$

Graph -

Theory used :

The area between the graph of $E_x$and the $x-$axis, where portions above the axis are taken with the sign and parts below the an axis are taken with the sign, is defined as minus the potential difference $∆V$.

Calculating the required the potential difference 

Starting at $200V/m$, the field component $E$declines for $100V/m$over a distance of $1m$, when$x=0$.

As a result, we can write

$$\begin{gathered} E_x=(200-\frac{100}{1m}(x-0\left)\right)V/m \\ \Rightarrow E_x=(200-100m^{-1}x)V/m \end{gathered}$$

The area between the graph of $E_x$and the $x-$axis, where portions above the axis are taken with the sign and parts below the an axis are taken with the sign, is defined as minus the potential difference $∆V$.

This area is then equal to the difference between the areas of the two right triangles in the diagram below:

That is,

$$\begin{gathered} \frac{1}{2}200V/m·1m-\frac{1}{2}100V/m·1m=0V \\ \Rightarrow \boxed{\mathbf{}\mathbf{∆}\mathit{V}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}\mathit{V}} \end{gathered}$$