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Calculate the ratio of the Coulomb electric force to the gravitational force between two electrons.

Physics

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Answer

4.164×10434.164 \times 10^{43}

Step-by-step solution

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To find the ratio of the Coulomb electric force (FEF_E) to the gravitational force (FGF_G) between two electrons in a vacuum, we will use Coulomb's Law and Newton's Law of Universal Gravitation.

Given constants:

  • Charge of an electron, e=1.602×10−19 Ce = 1.602 \times 10^{-19} \text{ C}
  • Mass of an electron, me=9.109×10−31 kgm_e = 9.109 \times 10^{-31} \text{ kg}
  • Coulomb's constant, k=8.987×109 N m2/C2k = 8.987 \times 10^9 \text{ N m}^2/\text{C}^2
  • Gravitational constant, G=6.674×10−11 N m2/kg2G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2

Let rr be the distance between the two electrons.

  1. Calculate the Coulomb electric force (FEF_E) between two electrons.

    The formula for the electric force between two point charges q1q_1 and q2q_2 separated by a distance rr is:

    FE=kq1q2r2F_E = \frac{k q_1 q_2}{r^2}

    For two electrons, q1=q2=eq_1 = q_2 = e.

    FE=ke2r2F_E = \frac{k e^2}{r^2}

  2. Calculate the gravitational force (FGF_G) between two electrons.

    The formula for the gravitational force between two masses m1m_1 and m2m_2 separated by a distance rr is:

    FG=Gm1m2r2F_G = \frac{G m_1 m_2}{r^2}

    For two electrons, m1=m2=mem_1 = m_2 = m_e.

    FG=Gme2r2F_G = \frac{G m_e^2}{r^2}

  3. Find the ratio FEFG\frac{F_E}{F_G}.

    FEFG=ke2r2Gme2r2\frac{F_E}{F_G} = \frac{\frac{k e^2}{r^2}}{\frac{G m_e^2}{r^2}}

    The r2r^2 terms cancel out:

    FEFG=ke2Gme2\frac{F_E}{F_G} = \frac{k e^2}{G m_e^2}

  4. Substitute the values and calculate the ratio.

    FEFG=(8.987×109 N m2/C2)(1.602×10−19 C)2(6.674×10−11 N m2/kg2)(9.109×10−31 kg)2\frac{F_E}{F_G} = \frac{(8.987 \times 10^9 \text{ N m}^2/\text{C}^2) (1.602 \times 10^{-19} \text{ C})^2}{(6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2) (9.109 \times 10^{-31} \text{ kg})^2} FEFG=(8.987×109)(2.566404×10−38)(6.674×10−11)(82.973881×10−62)\frac{F_E}{F_G} = \frac{(8.987 \times 10^9) (2.566404 \times 10^{-38})}{(6.674 \times 10^{-11}) (82.973881 \times 10^{-62})} FEFG=2.306×10−285.538×10−72\frac{F_E}{F_G} = \frac{2.306 \times 10^{-28}}{5.538 \times 10^{-72}} FEFG≈4.164×1043\frac{F_E}{F_G} \approx 4.164 \times 10^{43}

    The ratio of the Coulomb electric force to the gravitational force between two electrons in a vacuum is approximately 4.164×10434.164 \times 10^{43}.

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