4×10-10mWhat is the change in potential $\mathit{\Delta }\mathit{V}$in going from a location $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$ from a proton to a location$\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$ from the proton?

The initial location of the proton is,$r_i=3\times {10}^{-10}\text{m}$ .

The final location of the proton is, $r_f=4\times {10}^{-10}\text{m}$.

When an electric charge travels from one point to other then its electric potential changes between these two locations.

The change in the electric potential of the electric charge depends upon the travel distance of the charge as well as the magnitude of the electric charge.

The formula for the change in the potential from initial to final location is given by,

$$\begin{gathered} \Delta V=V_f-V_i \\ \Delta V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r_f}-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r_i} \\ \Delta V=\frac{1}{4\pi {\epsilon }_0}q\left(\frac{1}{r_f}-\frac{1}{r_i}\right) \end{gathered}$$

Here, $q=1.6\times {10}^{-19}\text{C}$is the charge of the proton, and the constant .

$\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\text{N}·{\text{m}}^2{\text{/C}}^2$.

Putting the values,

$$\begin{gathered} \Delta V=\left(9\times {10}^9\text{N}·{\text{m}}^2{\text{/C}}^2\right)\left(1.6\times {10}^{-19}\text{C}\right)\left(\frac{1}{4\times {10}^{-10}\text{m}}-\frac{1}{3\times {10}^{-10}\text{m}}\right) \\ \Delta V=14.4\times \left(\frac{1}{4}-\frac{1}{3}\right)\text{N}·\text{m/C}\times \frac{1\text{V}}{1\text{N}·\text{m/C}} \\ \Delta V=14.4\times \left(\frac{3-4}{12}\right)\text{V} \\ \Delta V=-1.2\text{V} \end{gathered}$$

The negative sign indicates that the potential is decreasing.

Hence, the change in potential is $-1.2\text{V}$.