In the Figure, location A is at ⟨ 0.5, 0, 0⟩ m, location C is ⟨1.3, 0, 0⟩ m, and location B is ⟨1.7, 0, 0⟩ m. ${\overrightarrow{\mathbf{E}}}_{\mathbf{1}}$= ⟨ 650, 0, 0⟩ N/C and ${\overrightarrow{\mathbf{E}}}_{\mathbf{2}}$=⟨-350, 0, 0⟩ N/C. Calculate the following quantities: (a)ΔV along a path going from A to B.

The given data can be listed below as,

• Location A, ${\mathrm{I}}_{\mathrm{A}}=0.5\mathrm{m}$
• Location B, ${\mathrm{I}}_{\mathrm{B}}=1.7\mathrm{m}$
• Location C, ${\mathrm{I}}_{\mathrm{c}}=1.3\mathrm{m}$
• Magnitude of the electric field A to B, ${\overrightarrow{\mathrm{E}}}_1=650\mathrm{V}/\mathrm{m}$
• Magnitude of the electric field B to A, ${\overrightarrow{\mathrm{E}}}_2=-350\mathrm{V}/\mathrm{m}$

The space diagram is as follows:

We divide the path into two segments by inventing a point C. Each segment of the path now passes through a region of uniform electric field.

Potential difference between any two points is defined as the amount of work done in moving a unit charge from one point to another. The potential difference between points A and B, VB-VA, is defined as the shift in the potential energy of a charge q, divided by the charge, shifted from A to B. Joules per coulomb, given the term volt (V) after Alessandro Volta, are units of potential difference.

The potential difference from pathAtoBis the summation of the potential differences of the pathACand the pathCB.

$∆\mathrm{V}=∆{\mathrm{V}}_{\mathrm{AC}}+∆{\mathrm{V}}_{\mathrm{CB}}$

The electric field is related to the potential difference between two points and the distance between them. Here, the potential difference (or the change in the electric potential) equals the displacement vector times the vector of the electric field

$∆\mathrm{V}=-\overrightarrow{\mathrm{E}}·∆\overrightarrow{\mathrm{I}}$

Hence, the potential difference for path AB is\

$$\begin{gathered} ∆\mathrm{V}=\left[-{\overrightarrow{\mathrm{E}}}_1\left({\mathrm{I}}_{\mathrm{C}}-{\mathrm{I}}_{\mathrm{A}}\right)\right]+\left[-{\overrightarrow{\mathrm{E}}}_2\left({\mathrm{I}}_{\mathrm{B}}-{\mathrm{I}}_{\mathrm{C}}\right)\right] \\ ∆\mathrm{V}=\left[-650\mathrm{V}/\mathrm{m}\left(1.3\mathrm{m}-0.5\mathrm{m}\right)\right]+\left[-\left(-350\mathrm{V}/\mathrm{m}\right)\left(1.7\mathrm{m}-1.3\mathrm{m}\right)\right] \\ ∆\mathrm{V}=-380\mathrm{V}/\mathrm{m} \end{gathered}$$