91Ó°ÊÓ

For charges${\mathit{q}}_{\mathbf{1}}$and${\mathit{q}}_{\mathbf{2}}$separated by a distance r, the magnitude of the electric force on eithercharge is proportional to the product${\mathit{q}}_{\mathbf{1}}{\mathit{q}}_{\mathbf{2}}$and inversely proportional to${\mathit{r}}^{\mathbf{2}}$.

$\mathit{F}\mathbf{=}\frac{\left|q_1{q}_2\right|}{{\mathbf{r}}^{\mathbf{2}}}$ …(1)

The rate at which an object's velocity changes with function of time is called acceleration.

The acceleration when the velocity is zero is called initial acceleration. The acceleration from the Newton’s second law is given by,

$\mathit{a}\mathbf{=}\frac{\mathbf{F}}{\mathbf{m}}$ …(2)

Here, mis the mass, and F is the force.

The fact that the charge of electron is equal to the charge of proton, which the equals to $1.60×{10}^{-19}\mathrm{C}$. The particles have an opposite charge, so they exert force on each other. By Coulomb’s law, the force is given by $F=\frac{\left|q_p{q}_e\right|}{r^2}$.

Substitute the value,$\frac{1}{4\mathrm{π}{\mathrm{ò}}_o}=9.0×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$ the charge,$q=1.60×{10}^{-19}\mathrm{C}$ and the distance,$r=2.0×{10}^{-10}\mathrm{m}$ in the above formula to get the force.

$$\begin{gathered} F=\frac{1}{4\mathrm{π}{\mathrm{ò}}_o}\frac{\left|q_p{q}_e\right|}{r^2} \\ =\left(9.0×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{{\left(1.60×{10}^{-19}\mathrm{C}\right)}^2}{{\left(2.0×{10}^{-10}\mathrm{m}\right)}^2} \\ =5.76×{10}^{-9}\mathrm{N} \end{gathered}$$

Substitute the force,$F=5.76×{10}^{-9}\mathrm{N}$ and the mass of the proton,$m_p=1.673×{10}^{27}\mathrm{kg}$ into the equation (2) to get the initial acceleration of the proton.

$$\begin{gathered} a_p=\frac{F}{m_p} \\ =\frac{5.76×{10}^{-9}\mathrm{N}}{1.673×{10}^{27}\mathrm{kg}} \\ =3.4×{10}^{18}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Substitute the force,role="math" localid="1668315377935" $F=5.76×{10}^{-9}N$ and the mass of the electron,$m_e=9.109×{10}^{-31}\mathrm{kg}$ into the equation (2) to get the initial acceleration of the electron.

$$\begin{gathered} a_e=\frac{F}{m_p} \\ =\frac{5.76×{10}^{-9}\mathrm{N}}{9.109×{10}^{-31}\mathrm{kg}} \\ =6.3×{10}^{21}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the initial acceleration of each particle is$3.4×{10}^{18}\mathrm{m}/{\mathrm{s}}^2$ and $6.3×{10}^{21}\mathrm{m}/{\mathrm{s}}^2$.