Question

You move from location i at $\mathbf{⟨}\mathbf{2}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{4}\mathbf{⟩}\mathbf{\hspace{0.33em}}\mathbf{m}$to location f at $\mathbf{⟨}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{9}\mathbf{⟩}\mathbf{\hspace{0.33em}}\mathbf{m}$. All along this path there is a nearby uniform electric field whose value is$\mathbf{⟨}\mathbf{1000}\mathbf{,}\mathbf{200}\mathbf{,}\mathbf{-}\mathbf{500}\mathbf{⟩}\mathbf{\hspace{0.33em}}\mathbf{N}\mathbf{/}\mathbf{C}\mathbf{.}\mathbf{}$ . Calculate,$\mathbf{\Delta V}\mathbf{=}{\mathbf{V}}_{\mathbf{f}}\mathbf{-}{\mathbf{V}}_{\mathbf{i}}$ including signs and units.

Short answer

The potential difference between two locations is$1500\hspace{0.33em}\mathrm{V}$

Step-by-step solution

Change in potential between two positions

The uniform electric field exerts a certain amount of force on the particle moving through the field between two locations.

The potential difference between two locations can be determined by the product of the uniform electric field and the distance between two locations.

Given data

The initial location is,$\overrightarrow{\mathrm{i}}=⟨2,5,4⟩\hspace{0.33em}\mathrm{m}$.

The final location is,$\overrightarrow{\mathrm{f}}=⟨3,5,9⟩\hspace{0.33em}\mathrm{m}.$

The value of the uniform electric field is, $\overrightarrow{\mathrm{E}}=⟨1000,200,-500⟩\hspace{0.33em}\mathrm{N}/\mathrm{C}$.

The potential difference between two locations

The distance moved from location i to location f is given by,

$$\begin{gathered} \mathrm{\Delta }{\overrightarrow{\mathrm{l}}}_{\mathrm{i}}\mathrm{f}=⟨3,5,9⟩\hspace{0.33em}\mathrm{m}-⟨2,5,4⟩\hspace{0.33em}\mathrm{m} \\ \mathrm{\Delta }{\overrightarrow{\mathrm{l}}}_{\mathrm{i}}\mathrm{f}=⟨1,0,5⟩\hspace{0.33em}\mathrm{m} \end{gathered}$$

Then the formula for the potential difference between two locations is given by,

Hence, the potential difference between two locations is $1500\hspace{0.33em}\mathrm{V}.$.