Question

(a)Two protons are a distance $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathbf{m}$apart. What is the electric potential energy of the system consisting of the two protons? If the two protons move closer together, will the electric potential energy of the system increase, decrease or remain the same? (b) A proton and an electron are a distance $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathbf{m}$apart. What is the electric potential energy of the system consisting of the proton and the electron? If the proton and the electron move closer together, will the electric potential energy of the system increase, decrease or remain the same? (c) Which of the following statements are true? A. In some situations, charged particles released from rest would move in a direction that increases electric potential energy, but not in other situations. B. If released from rest, two protons would move closer together, increasing the potential energy of the system. C. If any two charged particles are released from rest, they will spontaneously move in the direction in which the potential energy of the system will be decreased.

Short answer

1. The proton-proton system has the potential energy,${\mathit{U}}_{\mathbf{p}\mathbf{p}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{20}\mathbf{}\mathbf{}}\mathbf{J}$and the potential energy of the system will increase when the protons come closer.
2. The proton-electron system has the potential energy, ${\mathit{U}}_{\mathbf{pe}}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{20}\mathbf{}\mathbf{}}\mathbf{J}$and the potential energy of the system will decrease when the proton and electron will come closer.
3. The correct statement is C.

Step-by-step solution

Electric Potential Energy

The amount of work done in taking a positive charge from infinity to a particular point in the electric field of another charge gives the electric potential energy of the system of charges. When released, charged particles always move in the direction of decreasing potential energy. The electric potential energy$\left(U\right)$ of the system of two charges $\mathit{q}\mathbf{}\mathbf{and}\mathbf{}\mathbf{Q}$brought from infinity to separation is given as-
$\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\frac{\mathbf{q}\mathbf{Q}}{\mathbf{r}}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$
Here,${\mathbf{\epsilon }}_{\mathbf{o}}$is the permittivity of free space.

Given Data

Distance between two protons, $r=4\times {10}^{-9}\mathrm{m}$.

Distance between a proton and an electron, $r\text{'}=4\times {10}^{-9}\mathrm{m}$.

Part (a)

The charge on a proton is$q=+1.6\times {10}^{-19}\mathrm{C}$ and the separation of protons is $r=4\times {10}^{-9}\mathrm{m}$. Using equation (1), the potential energy of the proton-proton system $\left(U_{pp}\right)$is given as-

$U_{pp}=\frac{1}{4\pi {\epsilon }_o}\frac{q^2}{r}$

For the given values, the above equation becomes-

$$\begin{gathered} U_{pp}=\left(9\times {10}^9\raisebox{1ex}{${\mathrm{Nm}}^2$}\!\left/ \!\raisebox{-1ex}{${\mathrm{C}}^2$}\right.\right)\times \frac{{\left(1.6\times {10}^{-19}\mathrm{C}\right)}^2}{4\times {10}^{-9}\mathrm{m}} \\ =5.8\times {10}^{-20}\mathrm{J} \end{gathered}$$

From equation (1), the potential energy is inversely proportional to the separation of charges.

$U_{pp}\propto \frac{1}{r}$

So, when the protons move closer to each other, the separation $r$decreases and hence, the potential energy of the system increases.

Part (b)

The charge on a proton is $q=1.6\times {10}^{-19}\mathrm{C}$and on electron is$q\text{'}=-1.6\times {10}^{-19}\mathrm{C}$. The separation of proton and electron is $r\text{'}=4\times {10}^{-9}\mathrm{m}$. Using equation (1), the potential energy of the proton-electron system is given as-

role="math" localid="1661263159408" $U_{pe}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qq\text{'}}{r}$

For the given values, the above equation becomes-

role="math" localid="1661263677610" $$\begin{gathered} U_{pe}=(9\times {10}^9\raisebox{1ex}{${\mathrm{Nm}}^2$}\!\left/ \!\raisebox{-1ex}{${\mathrm{C}}^2$}\right.)\times \frac{(1.6\times {10}^{-19}\mathrm{C})(-1.6\times {10}^{-19}\mathrm{C})}{4\times {10}^{-9}\mathrm{m}} \\ =-5.8\times {10}^{-20}\mathrm{J} \end{gathered}$$

From equation (1), the potential energy is inversely proportional to the negative of separation of charges.

role="math" localid="1661263373469" $U_{pe}\propto -\frac{1}{r\text{'}}$

So, when the proton and electron move closer to each other, the separation$r\text{'}$ decreases and hence, the negative value of potential energy increases. This means that the potential energy of the proton-electron system decreases.

Part (c)

The correct statement is C.

When the two charged particles are released, they will spontaneously move in the direction in which the potential energy of the system will decrease.

Conclusion

1. The potential energy of the proton-proton system is $5.8\times {10}^{-20}\mathrm{J}$and the potential energy of system increases as the protons will come closer.
2. The potential energy of the proton-electron system is $-5.8\times {10}^{-20}\mathrm{J}$and the potential energy of system decreases as the proton and electron will come closer.
3. The correct statement is C.