You move from location $\mathit{i}$ at $⟨\mathbf{2}\mathbf{,}\mathbf{7}\mathbf{,}\mathbf{5}⟩\mathbf{m}$to location$\mathit{f}$ at $⟨\mathbf{5}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{12}⟩\mathbf{m}$. All along this path there is a nearly uniform electric field of $⟨\mathbf{1000}\mathbf{,}\mathbf{200}\mathbf{,}-\mathbf{510}⟩\mathbf{N}\mathbf{/}\mathbf{C}$. Calculate ${\mathbf{V}}_{\mathbf{f}}\mathbf{-}{\mathbf{V}}_{\mathbf{i}}$, including sign and units.

The given data can be listed below as,

• The coordinate of a location $i$ is,$\mathrm{i}⟨{\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1⟩=⟨2,7,5⟩\text{\hspace{0.17em}m}$ .
• The coordinate of another location $f$ is, $f⟨x_2,y_2,z_2⟩=⟨5,6,12⟩\text{\hspace{0.17em}m}$.
• The uniform electric field is, $⟨E_x,E_y,E_z⟩=⟨1000,200,-510⟩\text{\hspace{0.17em}N/C}$.

In this question, the electric field concept will be used to estimate the potential difference between two particular points in that electric field. The relation between the electric field and respective potential differences is direct linear.

The relation of potential difference between location $i$ and location$f$ is expressed as,

$\begin{array}{c}\mathrm{\Delta V}={\mathrm{V}}_{\mathrm{f}}-{\mathrm{V}}_{\mathrm{i}}\\ =-({\mathrm{E}}_{\mathrm{x}}\mathrm{\Delta x}+{\mathrm{E}}_{\mathrm{y}}\mathrm{\Delta y}+{\mathrm{E}}_{\mathrm{z}}\mathrm{\Delta z})\\ =-[{\mathrm{E}}_{\mathrm{x}}({\mathrm{x}}_2-{\mathrm{x}}_1)+{\mathrm{E}}_{\mathrm{y}}({\mathrm{y}}_2-{\mathrm{y}}_1)+{\mathrm{E}}_{\mathrm{z}}({\mathrm{z}}_2-{\mathrm{z}}_1)]\end{array}$

Here, $\mathrm{\Delta V}$ is the potential difference between two points.

Substitute all the known values in the above equation.

role="math" localid="1657121013389" $\begin{array}{c}{\mathrm{V}}_{\mathrm{f}}-{\mathrm{V}}_{\mathrm{i}}=-[(1000\mathrm{N}/\mathrm{C})(5\mathrm{m}-2\text{\hspace{0.17em}m})+\left(200\text{\hspace{0.17em}N/C}\right)(6\text{\hspace{0.17em}m}-7\text{\hspace{0.17em}m})+(-510\text{\hspace{0.17em}N/C})(12\text{\hspace{0.17em}m}-5\text{\hspace{0.17em}m})]\\ =-[-770\text{\hspace{0.17em}N}\cdot \text{m/C}]\\ =(770\text{\hspace{0.17em}N}\cdot \text{m/C})\times \left(\frac{1\text{\hspace{0.17em}V}}{1\text{\hspace{0.17em}N}\cdot \text{m/C}}\right)\\ =770\text{\hspace{0.17em}V}\end{array}$

Thus, the potential difference $({\mathrm{V}}_{\mathrm{f}}-{\mathrm{V}}_{\mathrm{i}})$ between location $i$ and$f$ location is $770\text{\hspace{0.17em}V}$.