Calculate the ratio of electrostatic to gravitational force between two electrons placed at certain distance in air. Given that $m_{e}$ = 9.1 x $10^{- 31}$ kg, e = 1.6 x $10^{- 19}$ C and $G = 6.6 x 10^{- 11} N m^{2} {kg}^{- 2}$

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$3.2 x 10^{41}$

$4.2 X 10^{42}$

$8.4 X 10^{42}$

$1.2 x 10^{42}$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{F_{elec}}{E_{gravi}} = \frac{1}{4 {πε}_{o}} . \frac{e \times e}{r^{2}} / G \frac{m_{e} \times m_{e}}{r^{2}} = \frac{1}{4 {πε}_{o}} . \frac{e^{2}}{{Gm}_{e}^{2}} \\ Putting G = 6 .6 \times 10^{- 11} {Nm}^{2} {kg}^{- 2}; \\ m_{e} = 9 .1 \times 10^{- 31} kg and e = 1 .6 \times 10^{- 19} C, we get \\ \frac{F_{elec}}{E_{gravi}} = 4 .2 \times 10^{42} \end{array}$