The Four protons, each with mass $\mathit{M}$and charge $\mathit{t}\mathit{e}$, are initially held at the corners of a square that is $\mathit{d}$on a side. They are then released from rest. What is the speed of each proton when the protons are very far apart?

Let us consider the figure,

The potential energy of a pair of charges separated by some distance is given by,

$U=K\frac{q_1{q}_2}{d}$

The electro static constant is given by $K;d$is the displacement between the charges and $q_1,q_2$are the magnitude of the charges.

The system given in the figure has four protons of charge $q$arranged to form four corners of a square.

So, the total initial potential energy of system is:

$U_{\mathrm{Te}}=U_{12}+U_{13}+U_{14}+U_{23}+U_{24}+U_{34}$..................(1)

Potential energy $U_{12}$is due to interaction of proton 1 and 2 , potential energy $U_{13}$is due to interaction of proton 1 and 3 , etc.

Now from the figure the displacement between protons $(1,2),(1,4),(2,3)and(3,4)$is$d$

Assume the charge on each of proton is equal and equal to $q$.

The potential energies of each pair is given by,

$$\begin{gathered} U_{12}=\frac{K{q}^2}{d} \\ {U}_{14}=\frac{K{q}^2}{d} \\ {U}_{23}-\frac{K{q}^2}{d} \\ {U}_{34}=\frac{K{q}^2}{d} \end{gathered}$$

Now, the displacement between protons $(2,4)and(1,3)$is $\sqrt{2}d$So,

$$\begin{gathered} U_{24}=\frac{K{q}^2}{\sqrt{2}d} \\ {U}_{13}=\frac{K{q}^2}{\sqrt{2}d} \end{gathered}$$

Next, substitute values of $U_{12},U_{13},U_{14},U_{23},U_{24}$and $U_{34}$in equation (1)

$$\begin{gathered} U_{T_0}=\frac{K{q}^2}{d}+\frac{K{q}^2}{d}+\frac{K{q}^2}{d}+\frac{K{q}^2}{d}+\frac{K{q}^2}{\sqrt{2}d}+\frac{K{q}^2}{\sqrt{2}d} \\ =4\frac{K{q}^2}{d}+\frac{2K{q}^2}{\sqrt{2}d} \\ =4\frac{K{q}^2}{d}+\frac{\sqrt{2}K{q}^2}{d} \end{gathered}$$

Also, if the protons are sufficiently far apart, the total potential energy of the system will be zero.

As a result of the conservation of energy principle, all of the original potential energy of the protons system will be turned into kinetic energy.

Therefore,

$K_{\text{Tes}}=U_{\text{Tet}}$

Replace $U_{ia}$ with $4\frac{K{q}^2}{d}+\frac{\sqrt{2}K{q}^2}{d}$

$K_{\mathrm{Te}}=4\frac{K{q}^2}{d}+\frac{\sqrt{2}K{q}^2}{d}$ ..............(2)

Again,

$K_{\mathrm{Ta}}=4\left(\frac{1}{2}m{v}^2\right)$

Here, $v$is the final velocity of proton and $m$is the mass of proton.

Substitute $K_{Ta}$by $\left(\frac{1}{2}m{v}^2\right)$in equation (2).

$$\begin{gathered} 4\left(\frac{1}{2}m{v}^2\right)=4\frac{K{q}^2}{d}+\frac{\sqrt{2}K{q}^2}{d} \\ {v}^2=\frac{2K{q}^2}{md}+\frac{K{q}^2}{\sqrt{2}md} \end{gathered}$$

Substitute $e$for $q$and $M$for$m$

$$\begin{gathered} v^2=\frac{2K{e}^2}{Md}+\frac{K{e}^2}{\sqrt{2}Md} \\ v={\left(\frac{2K{e}^2}{Md}+\frac{K{e}^2}{\sqrt{2Md}}\right)}^{\frac{1}{2}} \end{gathered}$$

Therefore, when two protons are separated by a large distance, their final velocity is

${\left(\frac{2K{e}^2}{Md}+\frac{K{e}^2}{\sqrt{2}Md}\right)}^{\frac{1}{2}}$