The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance $5 \times {10^{ - 11}}\,m,$ will be (Charge on electron $=$ $1.6 \times 10^{-19}$ $C$, mass of electron = $ 9.1 \times 10^{-31}$ $kg$, mass of proton = $1.6 \times {10^{ - 27}}\,kg,$ $\,G = 6.7 \times {10^{ - 11}}\,N{m^2}/k{g^2})$
The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance $5 \times {10^{ - 11}}\,m,$ will be (Charge on electron $=$ $1.6 \times 10^{-19}$ $C$, mass of electron = $ 9.1 \times 10^{-31}$ $kg$, mass of proton = $1.6 \times {10^{ - 27}}\,kg,$ $\,G = 6.7 \times {10^{ - 11}}\,N{m^2}/k{g^2})$
- A
$2.36 \times 10^{39}$
- B
$2.36 \times 10^{40}$
- C
$2.34 \times 10^{41}$
- D
$2.34 \times 10^{42}$
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$(ii)$ Suppose that the $\mathrm{Cs}$ atom at the corner $A$ is missing. What is the net force now on the $\mathrm{Cl}$ atom due to seven remaining $\mathrm{Cs}$ atoms ?
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