Electrostatics vs Gravitation!

Electricity and Magnetism Level 3

How many times is the electrostatic force of attraction , between two electrons separated by a distance of 'r' meters, greater than the gravitational 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 the Floor function

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

The answer is 46.

1 solution

Sravanth C.
Jan 10, 2016

We know that the electrostatics force is, $\vec{F_e}=\dfrac 1{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2} = \dfrac{9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{r^2}\\=\dfrac{2.56\times 9\times 10^{-29}}{r^2}$

And the force of gravitation is, $\vec{F_g}=\dfrac{Gm_1m_2}{r^2} = \dfrac{6\times10^{-11}\times 9\times 10^{-31}}{r^2}= \dfrac{486\times 10^{-73}}{r^2}$

$\therefore\dfrac{\vec{F_e}}{\vec{F_g}}=\dfrac{2.56\times 9\times 10^{-29}}{r^2}\times\dfrac{r^2}{486\times 10^{-73}} = 0.047407407\\= 4.740740\times 10^{-2}\times 10^{44} \boxed{\approx 4.7\times 10^{42}}$

So, $a = 4.7$ and $b = 42$ , hence $\lfloor 4.7 + 42 \rfloor = \boxed{46}$ .

Electrostatics vs Gravitation!

The answer is 46.

1 solution

0 pending reports