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State the formulas for electric and gravitational forces.

Physics

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Answer

2.2×1039\text{2.2} \times 10^{39}

Step-by-step solution

6 steps

To compare the electric and gravitational forces, we will calculate each force and then find their ratio.

  1. State the formulas for electric and gravitational forces.

    The electric force (FeF_e) between two charges q1q_1 and q2q_2 separated by a distance rr is given by Coulomb's Law:

    Fe=k∣q1q2∣r2F_e = k \frac{|q_1 q_2|}{r^2}

    The gravitational force (FgF_g) between two masses m1m_1 and m2m_2 separated by a distance rr is given by Newton's Law of Universal Gravitation:

    Fg=Gm1m2r2F_g = G \frac{m_1 m_2}{r^2}

  2. List the given values and physical constants.

    • Mass of electron, me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg}
    • Mass of proton, mp=1.7×10−27 kgm_p = 1.7 \times 10^{-27} \text{ kg}
    • Magnitude of charge of electron/proton, q=1.6×10−19 Cq = 1.6 \times 10^{-19} \text{ C}
    • Distance of separation, r=5.3×10−11 mr = 5.3 \times 10^{-11} \text{ m}
    • Coulomb's constant, k=8.99×109 N m2/C2k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2
    • Gravitational constant, G=6.67×10−11 N m2/kg2G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2

  3. Calculate the electric force (FeF_e).

    Using the formula Fe=kq2r2F_e = k \frac{q^2}{r^2}:

    Fe=(8.99×109 N m2/C2)(1.6×10−19 C)2(5.3×10−11 m)2F_e = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{(1.6 \times 10^{-19} \text{ C})^2}{(5.3 \times 10^{-11} \text{ m})^2} Fe=(8.99×109)2.56×10−3828.09×10−22F_e = (8.99 \times 10^9) \frac{2.56 \times 10^{-38}}{28.09 \times 10^{-22}} Fe=8.19×10−8 NF_e = 8.19 \times 10^{-8} \text{ N}

  4. Calculate the gravitational force (FgF_g).

    Using the formula Fg=Gmempr2F_g = G \frac{m_e m_p}{r^2}:

    Fg=(6.67×10−11 N m2/kg2)(9.1×10−31 kg)(1.7×10−27 kg)(5.3×10−11 m)2F_g = (6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \frac{(9.1 \times 10^{-31} \text{ kg})(1.7 \times 10^{-27} \text{ kg})}{(5.3 \times 10^{-11} \text{ m})^2} Fg=(6.67×10−11)15.47×10−5828.09×10−22F_g = (6.67 \times 10^{-11}) \frac{15.47 \times 10^{-58}}{28.09 \times 10^{-22}} Fg=3.67×10−47 NF_g = 3.67 \times 10^{-47} \text{ N}

  5. Determine the ratio of the electric force to the gravitational force.

    FeFg=8.19×10−8 N3.67×10−47 N\frac{F_e}{F_g} = \frac{8.19 \times 10^{-8} \text{ N}}{3.67 \times 10^{-47} \text{ N}} FeFg=2.23×1039\frac{F_e}{F_g} = 2.23 \times 10^{39}

  6. State the final comparison.

    The electric force between the electron and proton in a hydrogen atom is significantly stronger than the gravitational force.

    The ratio of the electric force to the gravitational force is 2.2×1039\boxed{\text{2.2} \times 10^{39}}.

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