Question

Check that the ratio $k{e}^2/G{m}_e{m}_p$ is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?

Answer 1

In the ratio $\frac{k{e}^2}{G{m}_e{m}_p},k=1/4\pi {\epsilon }_0\left(constant\right)$
Where, G=gravitational constant
$m_e$= mass of an electron
$m_p$= mass of a proton
From Coulomb's law
$F=k\frac{q_1{q}_2}{R^2}\Rightarrow k=\frac{F{r}^2}{q_1{q}_2}ork=\frac{F{r}^2}{q^2}$
The dimensions of $k=\left(\frac{1}{4\pi {\epsilon }_0}\right)=\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[AT\right]\left[AT\right]}=\left[M{L}^3{T}^{-4}{A}^{-2}\right]$
The dimensions of e(electronic charge)= [AT]
The dimensions of G (universal gravitational constant)
$\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{M^2}=\left[M^{-1}{L}^3{T}^{-2}\right]$
The dimensions of $m_e$ or $m_p$ (mass of electron or mass of proton)=[M]
The dimensions of $\frac{k{e}^2}{G{m}_e{m}_p}\frac{\left[M{L}^3{T}^{-4}{A}^{-2}\right]\left[A^2{T}^2\right]}{\left[M^{-1}{L}^3{T}^{-2}\right]\left[M^2\right]}=\left[M^0{L}^0{T}^0\right]$
Thus, the given ratio is dimensionless.
The value of $k=\left(\frac{1}{4\pi {\epsilon }_0}\right)=9\times {10}^{9}N-m^2/C^2$
The value of e (charge of an electron)= $1.6\times {10}^{-19}C$
The value of G(universal gravitational constant)= $6.67\times {10}^{-11}N-m^2/k{g}^2$
The value of $m_e$ (mass of proton)= $1.67\times {10}^{-27}kg$
The value of $\frac{k{e}^2}{G{m}_e{m}_p}=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{6.67\times {10}^{-11}\times 9.1\times {10}^{-31}\times 1.67\times {10}^{-27}}=2.29\times {10}^{39}$
The ratio signifies that the ratio of electrostatic force to the gravitational force is $2.29\times {10}^{39}.$ This means the electrostatic force between an electron and a proton is $2.29\times {10}^{39}$ times the gravitational force between an electron and a proton.