Q113 / 160 · JIPMER 2004 · Physics
The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance $5 \times 10^{11} \mathrm{~m}$, will be
(Charge on electron $=1.6 \times 10^{-19} \mathrm{C}$, mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$, mass of proton $=1.6 \times 10^{-27} \mathrm{~kg}$, $\left.G=6.7 \times 10^{-11} \quad \mathrm{Nm}^2 / \mathrm{kg}^2\right)$
- Correct answerA$2.36 \times 10^{39}$
- B$2.36 \times 10^{40}$
- C$2.34 \times 10^{41}$
- D$2.34 \times 10^{42}$
Solution
Gravitational force $F=\frac{G M_1 M_2}{r^2}$
$F_G=\frac{6.7 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-27}}{\left(5 \times 10^{-11}\right)^2}$
$=3.9 \times 10^{-47} \mathrm{~N}$
Electrostatic force $F_E=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$
$F_e=\frac{9 \times 10^9 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(5 \times 10^{-11}\right)^2}$
$=9.22 \times 10^{-8} \mathrm{~N}$
$\therefore \quad \frac{\text { Electrostatic force }}{\text { Gravitational force }}=\frac{F_E}{F_G}$
$\begin{aligned} & =\frac{9.22 \times 10^{-8}}{3.9 \times 10^{-47}} \\ & =2.36 \times 10^{39}\end{aligned}$
Hence, the correct answer is option: 'A'.