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Question: The ratio of electrostatic and gravitational forces acting between electron and proton separated by ...

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

A

2.36 × 1039

B

2.36 × 1040

C

2.34 × 1041

D

2.34 × 1042

Answer

2.36 × 1039

Explanation

Solution

Gravitational force FG=Gmempr2F_{G} = \frac{Gm_{e}m_{p}}{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}}{(5 \times 10^{- 11})^{2}}= 3.9 × 10–47 N

Electrostatic force Fe=14πε0e2r2F_{e} = \frac{1}{4\pi\varepsilon_{0}}\frac{e^{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}}{(5 \times 10^{- 11})^{2}} = 9.22 × 10–8 N

So, FeFG=9.22×10−83.9×10−47=2.36×1039\frac{F_{e}}{F_{G}} = \frac{9.22 \times 10^{- 8}}{3.9 \times 10^{- 47}} = 2.36 \times 10^{39}