Question

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?

Short answer

The change in potential is $-1.2\text{V}$.

Step-by-step solution

Identification of given data

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

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

Concept of electric potential

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.

Determining the change in potential

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 $-1.2\text{V}$.