91Ó°ÊÓ

Coulomb's law states that the magnitude F of the force exerted by two-point charges q1 and q2 separated by r is directly proportional to the product of the charges (q1*q2) and inversely proportional to the square of the distance between them.

$F=k\frac{\left|q_1{q}_2\right|}{r^2}$

Here;

F is the force on each point charge exerting on each other

K is the proportionality constant$k=8.98755×{10}^{9}N·m^2/C^2$

r is the distance between the charges

$q_1{q}_2$are the value of the two-point charges

Value given;

$$\begin{gathered} q_{proton}=1.60217×{10}^{-19}C \\ {m}_{proton}=1.6726×{10}^{-27}kg \end{gathered}$$

By Coulomb’s law;

$F=k\frac{\left|q_1{q}_2\right|}{r^2}$

Putting the values;

$$\begin{gathered} \mathrm{F}=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}=\mathrm{k}\frac{{{\mathrm{q}}_{\mathrm{proton}}}^2}{{\mathrm{r}}^2} \\ =\frac{\left(8.98755×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right){\left(1.60217×{10}^{-19}\mathrm{C}\right)}^2}{{\left(2.5×{10}^{-3}\mathrm{m}\right)}^2}=3.69×{10}^{-23}\mathrm{N} \end{gathered}$$

Using the force formula;

F=ma

By substitution

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}} \\ =\frac{3.69×{10}^{-23}\mathrm{N}}{1.6726×{10}^{-27}\mathrm{kg}} \\ =2.2×{10}^4\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the initial acceleration of the proton is$2.2×{10}^4\mathrm{m}/{\mathrm{s}}^2$