91Ó°ÊÓ

The given data can be listed below as,

• Potential difference, $∆\mathrm{V}$
• Initial Position, ${\mathrm{r}}_{\mathrm{i}}$
• Final Position, ${\mathrm{r}}_{\mathrm{f}}$
• Charge, $\mathrm{Q}$

The diagram is as follows:

The amount of labour required to move a unit charge from one point to another is the definition of potential difference between any two points. The change in the potential energy of a charge q divided by the charge shifted from point A to point B is known as the potential difference between points A and B, or VB-VA. Units of potential difference are called joules per coulomb, or volts (V) after Alessandro Volta.

The potential difference between two points is related to the charge Q by

$∆\mathrm{V}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\mathrm{Q}\left[\frac{1}{{\mathrm{r}}_{\mathrm{f}}}-\frac{1}{{\mathrm{r}}_{\mathrm{i}}}\right]$

For the closed path (1234), the potential difference across this path is the summation of the potential differences across each line $∆{\mathrm{V}}_1,∆{\mathrm{V}}_2,∆{\mathrm{V}}_3\mathrm{and}∆{\mathrm{V}}_4$

$∆V=∆{\mathrm{V}}_1+∆{\mathrm{V}}_2+∆{\mathrm{V}}_3+∆{\mathrm{V}}_4$

But, the electric field across line 2 and line 4 is perpendicular to each other, so the potential difference across these two lines is zero. Hence, the potential difference for the closed path is

$$\begin{gathered} ∆V=∆{\mathrm{V}}_1+∆{\mathrm{V}}_2+∆{\mathrm{V}}_3+∆{\mathrm{V}}_4 \\ ∆\mathrm{V}=∆{\mathrm{V}}_1+0+∆{\mathrm{V}}_3+0 \\ ∆\mathrm{V}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\mathrm{Q}\left[\frac{1}{{\mathrm{r}}_2}-\frac{1}{{\mathrm{r}}_1}\right]+\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\mathrm{Q}\left[\frac{1}{{\mathrm{r}}_1}-\frac{1}{{\mathrm{r}}_2}\right] \\ ∆\mathrm{V}=0 \end{gathered}$$