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NCERT Class XII Chapter
Electric Charges and Fields
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Question : 3 of 34
Marks: +1, -0
Check that the ratio Ke2Gmemp\frac{Ke^2}{G m_e m_p} is dimensionless. Look up a table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Solution:  
The ratio of electrostatic force to the gravitational force between an electron and a proton separated by a distance r from each other is
FelecFgrav\frac{F_{\text{elec}}}{F_{\text{grav}}} = Ke.r2Gmempr2\frac{\frac{Ke.}{r^2}}{\frac{G m_e m_p}{r^2}} = Ke2Gmemp\frac{Ke^2}{G m_e m_p}
In terms of dimensions
So, [Ke2Gmemp]\left[ \frac{Ke^2}{G m_e m_p} \right] = [FelecFgrav]\left[ \frac{F_{\text{elec}}}{F_{\text{grav}}} \right] = MLT2MLT2\frac{M L T^{-2}}{M L T^{-2}} = 1
Thus, the ratio Ke2Gmemp\frac{Ke^2}{G m_e m_p} is dimensionless
Ke2Gmemp\frac{Ke^2}{G m_e m_p} =
9×109×(1.6×1019)26.67×1011×9.1×1031Kg×1.67×1027Kg\frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \mathrm{Kg} \times 1.67 \times 10^{-27} \mathrm{Kg}}
= 0.23 × 104010^{40} = 2.3 × 103910^{39}
This ratio signifies that electrostatic forces are 103910^{39} times stronger than gravitational forces
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