Question

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.50 mm away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration–time and velocity–time graphs of the released proton’s motion.

Short answer

The initial acceleration of the proton is$2.2\times {10}^4\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Coulomb’s law

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$k=8.98755\times {10}^{9}N·m^2/C^2$

r is the distance between the charges

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

Initial acceleration of the proton

Value given;

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

By Coulomb’s law;

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

Putting the values;

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

Using the force formula;

F=ma

By substitution

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

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