Solveeit Logo
Login

Question

Physics Question on coulombs law

The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance 5×1011m5 \times 10^{11} m, will be (Charge on electron =1.6×10−19C=1.6 \times 10^{-19} C, mass of electron =9.1×10−31kg=9.1 \times 10^{-31} kg, mass of proton =1.6×10−27kg=1.6 \times 10^{-27} kg G=6.7×10−11Nm2/kg2)\left.G=6.7 \times 10^{-11} Nm ^{2} / kg ^{2}\right)

A

2.36×1039 2.36\times 10^{39}

B

2.36×1040 2.36\times 10^{40}

C

2.34×1041 2.34\times 10^{41}

D

2.34×1042 2.34\times 10^{42}

Answer

2.36×1039 2.36\times 10^{39}

Explanation

Solution

Gravitational force F=GM1M2r2F=\frac{G M_{1} M _{2}}{r^{2}}
FG=6.7×10−11×9.1×10−31×1.6×10−27(5×10−11)2F_{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}}
=3.9×10−47N=3.9 \times 10^{-47} N
Electrostatic force FE=14πε0q1q2r2F_{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}
Fe=9×109×1.6×10−19×1.6×10−19(5×10−11)2F_{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}}
=9.22×10−8N=9.22 \times 10^{-8} N
∴ Electrostatic force  Gravitational force =FEFG\therefore \frac{\text { Electrostatic force }}{\text { Gravitational force }}=\frac{F_{E}}{F_{G}}
=9.22×10−83.9×10−47=\frac{9.22 \times 10^{-8}}{3.9 \times 10^{-47}}
=2.36×1039=2.36 \times 10^{39}