Question

Two electrons are separated by one cm. What is the ratio of the electric force to the gravitational force between them? (me = 9. 11 ´ 10-31 kg, ke = 8. 99 ´ 109 N×m2 /C2 , G = 6. 67 ´ 10-11 N×m2 /kg2 , and e = 1. 6 ´ 10-19 C) a. 2. 3 ´ 102 b. 1. 3 ´ 1020 c. 3. 1 ´ 1022 d. 4. 2 ´ 1042.

Answer 1

The ratio of the electric force to the gravitational force between the two electrons is $4.2 \times 10^{42}$.

The given parameters;

• Mass of the electron, $m_e$= 9.11 x 10⁻¹¹ kg
• Coulomb's constant, k = 8.99 x 10⁹ Nm²/C²
• Gravitational constant, G = 6.67 x 10⁻¹¹ Nm²/kg²
• Charge of electron, q = 1.6 x 10⁻¹⁹ C
• Distance between the two electrons, r = 1 cm

The electrostatic force between the two electrons is calculated as follows;

$F_c = \frac{kq^2}{r^2}$

The gravitational force between the two electrons is calculated as follows;

$F_g = \frac{Gm_e^2}{r^2}$

The ratio of the electric force to the gravitational force between the two electrons is calculated as follows;

$\frac{F_c}{F_g} = \frac{kq^2}{r^2} \times \frac{1}{\frac{Gm^2}{r^2} } \\\\\frac{F_c}{F_g} = \frac{kq^2}{r^2} \times \frac{r^2}{Gm^2} \\\\\frac{F_c}{F_g} = \frac{kq^2}{Gm^2}$

$\frac{F_c}{F_g} = \frac{(8.99 \times 10^9) \times (1.6 \times 10^{-19}))^2}{(6.67 \times 10^{-11} ) \times (9.11 \times 10^{-31})^2} \\\\\frac{F_c}{F_g} = 4.2 \times 10^{42}$

Thus, the ratio of the electric force to the gravitational force between the two electrons is $4.2 \times 10^{42}$.

Learn more about electrostatic force here: https://brainly.com/question/16796365