Calculate the potential difference along the closed path consisting of two radial segments and two circular segments centred on the charge Q. Show that the four ΔV’s add up to zero. It is helpful to draw electric field vectors at several locations on each path segment to help keep track of signs.

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

$∆\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\mathrm{Q}\left[\frac{1}{{\mathrm{r}}_{\mathrm{f}}}-\frac{1}{{\mathrm{r}}_{\mathrm{i}}}\right]$

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

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