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Worked Examples · Example 1.3

Q.Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively.

(a) Compare the strength of these forces by determining the ratio of their magnitudes
(i) for an electron and a proton and
(ii) for two protons.
(b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 A˚ (=10−10 m)1\,\text{Å}\ (=10^{-10}\,\text{m}) apart? (mp=1.67×10−27 kgm_p = 1.67 \times 10^{-27}\,\text{kg}, me=9.11×10−31 kgm_e = 9.11 \times 10^{-31}\,\text{kg})
Telangana TsbieTextbookSubjective· 5mImportance★★★★★est
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The electrostatic force is immensely stronger than the gravitational force, by factors of about 2.3×10392.3 \times 10^{39} for an electron-proton pair and 1.2×10361.2 \times 10^{36} for two protons. At 1 A˚1\,\text{Å} separation, the electron experiences an acceleration of approximately 2.54×1022 m/s22.54 \times 10^{22}\,\text{m/s}^2, while the proton experiences an acceleration of about 1.38×1019 m/s21.38 \times 10^{19}\,\text{m/s}^2 due to their mutual electrical attraction.

Both Coulomb's law for electrostatic force and Newton's law for gravitational force describe interactions that follow an inverse-square dependence on the distance between the interacting particles. This means the force strength decreases rapidly as the separation increases. However, the fundamental nature and magnitudes of these forces are vastly different. Electrostatic force arises from charge and can be attractive or repulsive, while gravitational force arises from mass and is always attractive.

To truly grasp the relative importance of these forces, especially at the microscopic scale where atoms and subatomic particles interact, we compare their magnitudes directly. By forming a ratio of the electrostatic force to the gravitational force, we can quantify which force dominates. A key insight here is that since both forces depend on 1/r21/r^2, the distance rr will cancel out in the ratio. This means the relative strength of these fundamental forces is a constant, independent of how far apart the particles are.

(a) Comparing the strength of electrostatic and gravitational forces

  1. Formulate the force equations: The magnitude of the electrostatic force (FeF_e) between two point 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}

where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

The magnitude of the gravitational force ($F_g$) between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by Newton's law of gravitation:

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

where $G$ is the universal gravitational constant.

2. Determine the ratio of forces:

To compare their strengths, we take the ratio FeFg\frac{F_e}{F_g}:

FeFg=k∣q1q2∣r2Gm1m2r2\frac{F_e}{F_g} = \frac{k \frac{|q_1 q_2|}{r^2}}{G \frac{m_1 m_2}{r^2}}

As discussed, the $r^2$ terms cancel out, simplifying the ratio to:

FeFg=k∣q1q2∣Gm1m2\frac{F_e}{F_g} = \frac{k |q_1 q_2|}{G m_1 m_2}

  1. List fundamental constants:

    We use the following standard values for our calculations:

    • Elementary charge, e=1.602×10−19 Ce = 1.602 \times 10^{-19}\,\text{C}
    • Coulomb's constant, k=9×109 N m2/C2k = 9 \times 10^9\,\text{N m}^2/\text{C}^2
    • Universal gravitational constant, G=6.67×10−11 N m2/kg2G = 6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2
    • Mass of electron, me=9.11×10−31 kgm_e = 9.11 \times 10^{-31}\,\text{kg}
    • Mass of proton, mp=1.67×10−27 kgm_p = 1.67 \times 10^{-27}\,\text{kg}
    Watch out

    Precision in calculations involving powers of ten is crucial. Ensure correct substitution and arithmetic to avoid significant errors in the final magnitude.

    (i) For an electron and a proton:

    For an electron and a proton, the magnitude of their charges is ee for both, so ∣q1q2∣=e2|q_1 q_2| = e^2. Their masses are mem_e and mpm_p.

    Substituting these into the ratio formula:

FeFg=ke2Gmemp\frac{F_e}{F_g} = \frac{k e^2}{G m_e m_p}

FeFg=(9×109 N m2/C2)(1.602×10−19 C)2(6.67×10−11 N m2/kg2)(9.11×10−31 kg)(1.67×10−27 kg)\frac{F_e}{F_g} = \frac{(9 \times 10^9\,\text{N m}^2/\text{C}^2) (1.602 \times 10^{-19}\,\text{C})^2}{(6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2) (9.11 \times 10^{-31}\,\text{kg}) (1.67 \times 10^{-27}\,\text{kg})}

Calculating the numerator:

(9×109)×(1.602)2×10−38=9×2.566404×10−29≈2.310×10−28(9 \times 10^9) \times (1.602)^2 \times 10^{-38} = 9 \times 2.566404 \times 10^{-29} \approx 2.310 \times 10^{-28}

Calculating the denominator:

(6.67×10−11)×(9.11×10−31)×(1.67×10−27)=(6.67×9.11×1.67)×10−11−31−27(6.67 \times 10^{-11}) \times (9.11 \times 10^{-31}) \times (1.67 \times 10^{-27}) = (6.67 \times 9.11 \times 1.67) \times 10^{-11-31-27}

=101.399×10−69≈1.014×10−67= 101.399 \times 10^{-69} \approx 1.014 \times 10^{-67}

Now, divide the numerator by the denominator:

FeFg=2.310×10−281.014×10−67≈2.278×1039\frac{F_e}{F_g} = \frac{2.310 \times 10^{-28}}{1.014 \times 10^{-67}} \approx 2.278 \times 10^{39}

Rounding to two significant figures, consistent with the precision of $k$ and $G$:

FeFg≈2.3×1039\frac{F_e}{F_g} \approx \mathbf{2.3 \times 10^{39}}

This result highlights that the electrostatic force is astronomically stronger than the gravitational force at the atomic scale.

**(ii) For two protons:**
For two protons, the magnitude of their charges is $e$ for both, so $|q_1 q_2| = e^2$. Their masses are both $m_p$.
Substituting these into the ratio formula:

FeFg=ke2Gmp2\frac{F_e}{F_g} = \frac{k e^2}{G m_p^2}

FeFg=(9×109 N m2/C2)(1.602×10−19 C)2(6.67×10−11 N m2/kg2)(1.67×10−27 kg)2\frac{F_e}{F_g} = \frac{(9 \times 10^9\,\text{N m}^2/\text{C}^2) (1.602 \times 10^{-19}\,\text{C})^2}{(6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2) (1.67 \times 10^{-27}\,\text{kg})^2}

The numerator is the same as before: $2.310 \times 10^{-28}$.
Calculating the denominator:

(6.67×10−11)×(1.67×10−27)2=(6.67×(1.67)2)×10−11−54(6.67 \times 10^{-11}) \times (1.67 \times 10^{-27})^2 = (6.67 \times (1.67)^2) \times 10^{-11-54}

=(6.67×2.7889)×10−65=18.598×10−65≈1.860×10−64= (6.67 \times 2.7889) \times 10^{-65} = 18.598 \times 10^{-65} \approx 1.860 \times 10^{-64}

Now, divide the numerator by the denominator:

FeFg=2.310×10−281.860×10−64≈1.242×1036\frac{F_e}{F_g} = \frac{2.310 \times 10^{-28}}{1.860 \times 10^{-64}} \approx 1.242 \times 10^{36}

Rounding to two significant figures:

FeFg≈1.2×1036\frac{F_e}{F_g} \approx \mathbf{1.2 \times 10^{36}}

Even for two protons, the electrostatic repulsion is vastly stronger than their gravitational attraction. The ratio is smaller than for the electron-proton pair because protons are much more massive than electrons, leading to a larger gravitational force in the denominator.

(b) Estimating accelerations

  1. Understand acceleration from force: According to Newton's second law, the acceleration (aa) of an object is directly proportional to the net force (FF) acting on it and inversely proportional to its mass (mm):

a=Fma = \frac{F}{m}

We need to calculate the acceleration of both the electron and the proton due to their mutual electrical attraction.

2. Calculate the electrostatic force:

The electron and proton are separated by a distance r=1 A˚=10−10 mr = 1\,\text{Å} = 10^{-10}\,\text{m}.

The magnitude of the electrostatic force between them is:

Fe=ke2r2F_e = k \frac{e^2}{r^2}

Fe=(9×109 N m2/C2)(1.602×10−19 C)2(10−10 m)2F_e = (9 \times 10^9\,\text{N m}^2/\text{C}^2) \frac{(1.602 \times 10^{-19}\,\text{C})^2}{(10^{-10}\,\text{m})^2}

Fe=(9×109)2.566404×10−3810−20F_e = (9 \times 10^9) \frac{2.566404 \times 10^{-38}}{10^{-20}}

Fe=9×2.566404×109−38+20F_e = 9 \times 2.566404 \times 10^{9 - 38 + 20}

Fe=23.097636×10−9 NF_e = 23.097636 \times 10^{-9}\,\text{N}

Fe≈2.31×10−8 NF_e \approx \mathbf{2.31 \times 10^{-8}\,\text{N}}

This is the magnitude of the attractive force acting on both the electron and the proton.

3. Calculate the acceleration of the electron:

The force on the electron is FeF_e, and its mass is me=9.11×10−31 kgm_e = 9.11 \times 10^{-31}\,\text{kg}.

ae=Femea_e = \frac{F_e}{m_e}

ae=2.30976×10−8 N9.11×10−31 kga_e = \frac{2.30976 \times 10^{-8}\,\text{N}}{9.11 \times 10^{-31}\,\text{kg}}

ae≈0.2535×1023 m/s2a_e \approx 0.2535 \times 10^{23}\,\text{m/s}^2

ae≈2.54×1022 m/s2a_e \approx \mathbf{2.54 \times 10^{22}\,\text{m/s}^2}

  1. Calculate the acceleration of the proton: The force on the proton is also FeF_e (by Newton's third law), and its mass is mp=1.67×10−27 kgm_p = 1.67 \times 10^{-27}\,\text{kg}.

ap=Fempa_p = \frac{F_e}{m_p}

ap=2.30976×10−8 N1.67×10−27 kga_p = \frac{2.30976 \times 10^{-8}\,\text{N}}{1.67 \times 10^{-27}\,\text{kg}}

ap≈1.383×1019 m/s2a_p \approx 1.383 \times 10^{19}\,\text{m/s}^2

ap≈1.38×1019 m/s2a_p \approx \mathbf{1.38 \times 10^{19}\,\text{m/s}^2}

> [!TIP]
> The electron, being significantly less massive than the proton ($m_p \approx 1836 m_e$), experiences a proportionally larger acceleration for the same magnitude of force. This is a direct consequence of Newton's second law, $a = F/m$.
✓Final answer

  1. The ratio of electrostatic to gravitational force is approximately 2.3×1039\boxed{2.3 \times 10^{39}} for an electron and a proton, and 1.2×1036\boxed{1.2 \times 10^{36}} for two protons.
  2. The acceleration of the electron is approximately 2.54×1022 m/s2\boxed{2.54 \times 10^{22}\,\text{m/s}^2}, and the acceleration of the proton is approximately 1.38×1019 m/s2\boxed{1.38 \times 10^{19}\,\text{m/s}^2}.

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