Question

Compare the magnitude of the electromagnetic and gravitational force between two electrons separated by a distance of 2.00 m. Assume the electrons have a mass of 9.11 × 10^–31 kg and a charge of 1.61 × 10^–19 C. Round to two decimal places.
Fe = _____ × 10^–29 N
Fg = _____ × 10^–71 N
Fe/ Fg= _____ × 10^42

Please answer as soon as possible!

Answer 1

First you need to know about two laws, which are:
1) Coulomb's law
2) Newton's law of gravitation

1.
According to Coulomb's law, Electric force between TWO charges is:
$F_{e} = \frac{k*q_{1}*q_{2}}{r^{2}}$ -- (A)

Where, k = 1/(4*π*epsilon_not) = $9 * 10^{9}$ $\frac{Nm^{2}}{C^{2}}$

Both $q_{1}$ and $q_{2}$ = -1.61 x $10^{-19}$ C
r = Distance between the two charges = 2.00m

Plug-in the above values in (A), you would get:

$F_{e}$ = (9 * $10^{9}$ ) (1.61 * $10^{-19}$  * 1.61 * $10^{-19}$) / (2*2)  

$F_{e}$ = $5.83* 10^{-29}$ N

2.
According to Newton's law of gravitation:
$F_{g} = \frac{Gm_{1}m_{2}}{r^{2}}$ -- (B)

Where G = Gravitational constant = 6.674 * $10^{-11} m^{2} kg^{-1}s^{-2}$

m1 = m2 = Mass of the electron = $9.11 * 10^{-31}$ kg
r = 2.0 m

Plug-in the above values in (B), you would get:

$F_{g}$ = (6.674 * $10^{-11}$) ( $9.11 * 10^{-31}$  *  $9.11 * 10^{-31}$}[/tex]) / (2*2)  

$F_{e}$ = $5.83* 10^{-29}$ N

$F_{e}$ = $1.38 * 10^{-71} N$

Now do Fe over Fg, you would get:
$\frac{F_{e}}{F{g}} = 4.23 * 10^{42}$

Ans: So the blanks are:
1) 5.82
2) 1.38
3) 4.23

-i

Answer 2

1) 5.83

2) 1.38

3) 4.22