Question

$F_{g}$ and $F_{e}$ represents gravitational and electrostatic force respectively between electron and proton at a distance of $10\, cm$. The ratio of $F_{g} / F_{e}$ is of the order of

• A. $10^{42}$
• B. $10^{-39}$
• C. $1$
• D. $10^{-43}$

Answer 1

Answer: (B)

The electric force between an electron and proton at a distance $r$ apart is
$F_{e}=\frac{-k e^{2}}{r^{2}}\,\,\,...(i)$
where, the negative sign indicates that the force is attractive. The corresponding gravitational force (always attractive) is
$F_{G}=-G \frac{m_{p} m_{e}}{r^{2}}\,\,\,\,...(ii)$
where, $m_{p}$ and $m_{e}$ are the masses of the proton and electron.
On comparing Eqs. (i) and (ii), we get
$\left|\frac{F_{e}}{F_{G}}\right|=\frac{k e^{2}}{G m_{p} m_{e}}=2.4 \times 10^{39}$
$\Rightarrow F_{G} / F_{e} \approx 10^{-39}$