91Ó°ÊÓ

The given data can be listed below as:

• The location of the proton is$(0,0,-2×{10}^{-9}) \mathrm{m}$ .
• The location of the alpha particle is$(1.5×{10}^{-9},0,2×{10}^{-9}) m$ .

This law states that the unlike charges attracts and the like charges also repel each other. The electrostatic force is directly proportional to the product of the charges and inversely proportional to the square of their distances.

The equation of the electrostatic force gives the force exerted by the proton and also the alpha particle.

The location of the proton can be expressed as:

$r_1=\left(-2×{10}^{-9}\mathrm{m}\right)\stackrel{→}{k}$

The location of the electron can be expressed as:

$r_2=\left(1.5×{10}^{-9}\mathrm{m}\right)\stackrel{→}{i}+\left(2×{10}^{-9}\mathrm{m}\right)\stackrel{→}{k}$

Hence, the position vector can be expressed as:

$$\begin{gathered} r=r_1-r_2 \\ r=\left(1.5×{10}^{-9}\mathrm{m}\right)\stackrel{→}{i}+\left(2×{10}^{-9}\mathrm{m}\right)\stackrel{→}{k}-\left(-2×{10}^{-9}\mathrm{m}\right)\stackrel{→}{k} \\ r=\left(1.5×{10}^{-9}\mathrm{m}\right)\stackrel{→}{i}+\left(4×{10}^{-9}\mathrm{m}\right)\stackrel{→}{k} \end{gathered}$$

The distance of the particle is expressed as:

$$\begin{gathered} r=\sqrt{{\left(1.5×{10}^{-9}\mathrm{m}\right)}^2+{\left(4×{10}^{-9}\mathrm{m}\right)}^2} \\ r=4.272×{10}^{-9}\mathrm{m} \end{gathered}$$

a) From the coulomb’s law, the force exerted by the proton on the alpha particle is expressed as:

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

Here, F is the force exerted by the proton, $q_{1}and{q}_2$is the charges of the proton and the alpha particle which are$1.6022×{10}^{-19}Cand3.2×{10}^{-19}C$respectively, k is the coulomb’s constant that is localid="1658113098228" $9.0×{10}^9\mathrm{N}.m^2/{\mathrm{C}}^2$and r is the distance amongst the particles that is $4.272×{10}^{-9}\mathrm{m}$.

Substituting the values in the above equation, we get-

$$\begin{gathered} \mathrm{F}=9.0×{10}^9{\mathrm{N}.\mathrm{m}}^2/{\mathrm{C}}^2×\frac{\left(1.6022×{10}^{-19}\mathrm{C}\right)×\left(3.2×{10}^{-19}\mathrm{C}\right)}{{\left(4.272×{10}^{-9}\mathrm{m}\right)}^2} \\ \mathrm{F}=2.528×{10}^{-11}\mathrm{N} \end{gathered}$$

Thus, the force the proton exerts on the alpha particle islocalid="1658112897340" $2.528×{10}^{-11}\mathrm{N}$

b)From the coulomb’s law, the force exerted by the alpha particle on the proton is expressed as:

$F=k\frac{q_2{q}_1}{r^2}$

Here, F is the force exerted by the proton $q_1\text{and}{q}_2$, is the charges of the proton and the alpha particle which are$1.6022×{10}^{-19}\mathrm{C}\text{and}3.2×{10}^{-19}$respectively, k is the coulomb’s constant that is$9.0×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$and r is the distance amongst the particles that is$4.272×{10}^{-9} m$.

Substituting the values in the above equation, we get-

localid="1658112939654" $$\begin{gathered} F=9.0×{10}^9{\mathrm{N}.\mathrm{m}}^2/{\mathrm{C}}^2×\frac{\left(1.6022×{10}^{-19}\mathrm{C}\right)×\left(3.2×{10}^{-19}\mathrm{C}\right)}{{\left(4.272×{10}^{-9}\mathrm{m}\right)}^2} \\ F=2.528×{10}^{-11}\mathrm{N} \end{gathered}$$

Thus, the force the alpha particle exerts on the proton is $2.528×{10}^{-11}\mathrm{N}$