Electrostatics vs Gravitation!

是多少次electrostatic force of attraction, between two electrons separated by a distance of 'r' meters, greater than thegravitational force of attraction?

The answer is in the form $a\times10^b$(in standard form), provide the answer as $\lfloor a+b\rfloor$.

Details and Assumptions:

• Charge on an electron $=1.6\times 10^{-19}\text{C}$,

• Mass of electron $= 9\times 10^{-31}\text{Kg}$

• Gravitational Constant $=6\times 10^{-11}\text{Nm}^2\text{/kg}^2$and

• $\dfrac 1{4\pi\epsilon_0}=9\times 10^9\text{Nm}^2\text{/C}^2$.

• $\lfloor.\rfloor$is theFloor function

Note:The values given above are approximations, do not use them as specific values.