If a proton and an electron are released when they are 2.0 \times 10^{- 10}m

Step 1
Given
We are given the distance between a proton and an electron ${\displaystyle r=2.0\times {10}^{-10}m}$. We are asked to calculate initial acceleration of each particle.
Solution
The two particles have an opposite charge, therefore each one exerts a force F on the other particle. This force could be calculated by the Coulomb's law as in equalion 21.2
$F=\frac{1}{4\pi {ϵ}_0}\frac{\mid q_p{q}_e\mid }{r^2}$
Where ${\displaystyle {ϵ}_0}$ is the electric constant and the term ${\displaystyle \frac{1}{4\pi {ϵ}_0}}$ equals ${\displaystyle 9.0\times {10}^{9}N\cdot \frac{m^2}{C^2}}$. ${\displaystyle q_p}$ is the charge of proton and equals ${\displaystyle 1.60\times {10}^{-19}C}$ and ${\displaystyle q_e}$, is the charge of electron and equals ${\displaystyle 1.60\times {10}^{-19}C}$ (See Appendix F)
To obtain the initial acceleration a of the two particles we should get the force that exerted on the proton and the electron. Where the acceleration a is given by Newton's second law in the form.
${\displaystyle a=\frac{F}{m}}$
Where m is the mass of the particle, and for proton ${\displaystyle m_p=1.67\times {10}^{-19}kg}$ while for electron ${\displaystyle m_e=9.11\times {10}^{-31}kg}$ (See Appendix F)
Now we can plug our values tor ${\displaystyle g_p}$, ${\displaystyle g_e}$ and ${\displaystyle \tau }$ into equation (1) to get F
$F=\frac{1}{4\pi {ϵ}_0}\frac{\mid q_p{q}_e\mid }{r^2}$
${\displaystyle =(9.0\times {10}^{9}N\cdot \frac{m^2}{C^2})\frac{{(1.60\times {10}^{-19}C)}^2}{{(2.0\times {10}^{-10}m)}^2}}$
${\displaystyle =5.76\times {10}^{-9}N}$
Step 2
Now, use equation (2) to get the acceleration a for each particle using the value of F.
For proton: We can plug our values for F and ${\displaystyle m_p}$ into equation (2) to get the initial acceleration a of the proton
${\displaystyle a_p=\frac{F}{m_p}=\frac{5.76\times {10}^{-9}N}{1.673\times {10}^{-27}kg}=3.4\times {10}^{18}\frac{m}{s^2}}$
For electron Again plug our values for F and ${\displaystyle m_n}$, into equation (2) to gel the initial acceleration a of the electron
${\displaystyle a_e=\frac{F}{m_p}=\frac{5.76\times {10}^{-9}N}{9.109\times {10}^{-31}kg}=6.3\times {10}^{21}\frac{m}{s^2}}$

Proton
${\displaystyle m_p=1.63\times {10}^{-27}kg}$
${\displaystyle q_p=e=1.602\times {10}^{-19}C}$
Electron
${\displaystyle m_p=9.109\times {10}^{-31}kg}$
${\displaystyle q_p=-e=-1.602\times {10}^{-19}C}$
Since the charges are opposite, the Coulon b fuse is attractive.
$F=\frac{1}{4\pi {ϵ}_0}\frac{\mid q_p{q}_e\mid }{r^2}=(8.988\times {10}^9\frac{N{m}^2}{c^2})\frac{{(1.602\times {10}^{-19}C)}^2}{{(2.0\times {10}^{-10}m)}^2}=5.766709788\times {10}^{-9}$
From Newtois Second Law, ${\displaystyle F=ma\Rightarrow a=\frac{F}{m}}$
${\displaystyle \therefore }$ acceleration of Proton:
${\displaystyle a_p=\frac{F}{m_p}=\frac{5.766709788\times {10}^{-9}N}{1.673\times {10}^{-27}kg}=3.4469\times {10}^{18}\frac{m}{v^2}}$
${\displaystyle \therefore a_p=3.4\times {10}^{18}\frac{m}{v^2}}$
acceleration of electron:
${\displaystyle a_e=\frac{F}{m_e}=\frac{5.766709788\times {10}^{-9}N}{9.109\times {10}^{-31}kg}=6.33078\times {10}^{27}\frac{m}{v^2}}$
${\displaystyle \therefore a_p=6.3\times {10}^{18}\frac{m}{v^2}}$