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

\(F=\frac{1}{4\pi\epsilon_{0}}\frac{\mid q_{p}q_{e}\mid}{r^{2}}\)

Where \(\displaystyle\epsilon_{{{0}}}\) is the electric constant and the term \(\displaystyle{\frac{{{1}}}{{{4}\pi\epsilon_{{{0}}}}}}\) 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 \(\displaystyle{m}_{{{p}}}={1.67}\times{10}^{{-{19}}}{k}{g}\) while for electron \(\displaystyle{m}_{{{e}}}={9.11}\times{10}^{{-{31}}}{k}{g}\) (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

\(F=\frac{1}{4\pi\epsilon_{0}}\frac{\mid q_{p}q_{e}\mid}{r^{2}}\)

\(\displaystyle={\left({9.0}\times{10}^{{{9}}}{N}\cdot\frac{{m}^{{{2}}}}{{C}^{{{2}}}}\right)}{\frac{{{\left({1.60}\times{10}^{{-{19}}}{C}\right)}^{{{2}}}}}{{{\left({2.0}\times{10}^{{-{10}}}{m}\right)}^{{{2}}}}}}\)

\(\displaystyle={5.76}\times{10}^{{-{9}}}{N}\)

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

\(\displaystyle{a}_{{{p}}}={\frac{{{F}}}{{{m}_{{{p}}}}}}={\frac{{{5.76}\times{10}^{{-{9}}}{N}}}{{{1.673}\times{10}^{{-{27}}}{k}{g}}}}={3.4}\times{10}^{{{18}}}\frac{{m}}{{s}^{{{2}}}}\)

For electron Again plug our values for F and \(\displaystyle{m}_{{{n}}}\), into equation (2) to gel the initial acceleration a of the electron

\(\displaystyle{a}_{{{e}}}={\frac{{{F}}}{{{m}_{{{p}}}}}}={\frac{{{5.76}\times{10}^{{-{9}}}{N}}}{{{9.109}\times{10}^{{-{31}}}{k}{g}}}}={6.3}\times{10}^{{{21}}}\frac{{m}}{{s}^{{{2}}}}\)