Question

Figure EX26.9shows a graph of $V\mathrm{versus}x$in a region of space. The potential is independent of y and z. What is $E_x$at

(a) $x=-2cm$, (b) $x=0cm$, and (c) $x=2cm$?

Short answer

a.$E_{x=-2}=\mathbf{5}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}$

b.$E_{x=0cm}=\mathbf{-}\mathbf{10}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}$

c.$E_{x=2cm}=\mathbf{5}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}$

Step-by-step solution

Given information and formula used   

Given graph :

Positions : (a) $x=-2cm$, (b) $x=0cm$, and (c) $x=2cm$

Theory used :

The electric field strength is assumed to be independent of both $y\mathrm{and}z$positions, i.e., it only depends on $x$position. As a result,

$E=-\frac{dV}{dx}$

Calculating Ex at (a) x = -2 cm, (b) x = 0 cm, and (c) x = 2 cm

Finding the derivative of the potential at the stated points of interest, we have :

a. At a distance of $x=-2cm$, the electric field strength will be :

$$\begin{gathered} E_{x=-2cm}=-\frac{d}{dx}{V}_{x=-2cm} \\ =-\frac{-10-0}{-1-(-3)}=\boxed{\mathbf{5}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}} \end{gathered}$$

b. At position $x=0cm$, the electric field strength will be

$$\begin{gathered} E_{x=0cm}=-\frac{d}{dx}{V}_{x=0cm} \\ =-\frac{10-(-10)}{1-(-1)}=\boxed{\mathbf{-}\mathbf{10}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}} \end{gathered}$$

c. At location $x=2cm$, the electric field strength will be

$$\begin{gathered} E_{x=2cm}=-\frac{d}{dx}{V}_{x=2cm} \\ =-\frac{0-10}{3-1}=\boxed{\mathbf{5}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}} \end{gathered}$$