Coulomb's Law vs Gravitational Force — Why Electrostatic Is Stronger

easy CBSE JEE-MAIN NCERT Class 12 4 min read

Question

Compare the electrostatic force and gravitational force between a proton and an electron. Show that the electrostatic force is far stronger, and calculate the ratio Fe/FgF_e / F_g.

Given: mass of proton mp=1.67×1027m_p = 1.67 \times 10^{-27} kg, mass of electron me=9.11×1031m_e = 9.11 \times 10^{-31} kg, charge on each =1.6×1019= 1.6 \times 10^{-19} C, k=9×109k = 9 \times 10^9 N m² C⁻², G=6.67×1011G = 6.67 \times 10^{-11} N m² kg⁻².


Solution — Step by Step

Both forces follow the inverse square law, so the r2r^2 terms will cancel when we take the ratio — which is why we don’t need to know the actual separation distance.

Fe=ke2r2,Fg=Gmpmer2F_e = k \frac{e^2}{r^2}, \quad F_g = G \frac{m_p \, m_e}{r^2}

Divide FeF_e by FgF_g. The r2r^2 cancels cleanly:

FeFg=ke2Gmpme\frac{F_e}{F_g} = \frac{k \, e^2}{G \, m_p \, m_e}

This ratio is a pure number — same regardless of where the proton and electron are sitting.

ke2=9×109×(1.6×1019)2k \, e^2 = 9 \times 10^9 \times (1.6 \times 10^{-19})^2 =9×109×2.56×1038= 9 \times 10^9 \times 2.56 \times 10^{-38} =23.04×1029=2.304×1028 N m2= 23.04 \times 10^{-29} = 2.304 \times 10^{-28} \text{ N m}^2 Gmpme=6.67×1011×1.67×1027×9.11×1031G \, m_p \, m_e = 6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.11 \times 10^{-31}

Work step by step: 1.67×9.1115.211.67 \times 9.11 \approx 15.21, so mpme15.21×1058=1.521×1057m_p m_e \approx 15.21 \times 10^{-58} = 1.521 \times 10^{-57} kg²

Gmpme=6.67×1011×1.521×1057=10.15×10681.015×1067 N m2G \, m_p \, m_e = 6.67 \times 10^{-11} \times 1.521 \times 10^{-57} = 10.15 \times 10^{-68} \approx 1.015 \times 10^{-67} \text{ N m}^2
FeFg=2.304×10281.015×10672.27×1039\frac{F_e}{F_g} = \frac{2.304 \times 10^{-28}}{1.015 \times 10^{-67}} \approx 2.27 \times 10^{39}

The electrostatic force is approximately 2.27×10392.27 \times 10^{39} times stronger than the gravitational force between a proton and electron.


Why This Works

Both Coulomb’s law and Newton’s law of gravitation are inverse square laws — the force drops as 1/r21/r^2 in both cases. When you form the ratio, the distance rr completely cancels. This is why the answer is universal: it doesn’t matter if the proton and electron are 1 nm apart or 1 km apart, the ratio stays the same.

The enormous factor of 103910^{39} comes from the constants themselves. The Coulomb constant k9×109k \approx 9 \times 10^9 is huge, while G6.67×1011G \approx 6.67 \times 10^{-11} is tiny. Electric charges (1019\sim 10^{-19} C) are small but their square is still far larger than the product of electron and proton masses (1057\sim 10^{-57} kg²) when scaled by their respective constants.

This result has deep physical meaning: gravity is completely negligible at the atomic and molecular scale. Atoms, chemical bonds, and everyday materials are held together by electromagnetic forces — gravity simply doesn’t participate at that level.


Alternative Method

Instead of computing the ratio symbolically, you can compute each force separately at a fixed separation (say r=1r = 1 Å =1010= 10^{-10} m) and then divide.

Fe=9×109×(1.6×1019)2(1010)2=2.304×10281020=2.304×108 NF_e = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(10^{-10})^2} = \frac{2.304 \times 10^{-28}}{10^{-20}} = 2.304 \times 10^{-8} \text{ N} Fg=6.67×1011×1.67×1027×9.11×1031(1010)2=1.015×106710201.015×1047 NF_g = \frac{6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.11 \times 10^{-31}}{(10^{-10})^2} = \frac{1.015 \times 10^{-67}}{10^{-20}} \approx 1.015 \times 10^{-47} \text{ N} FeFg=2.304×1081.015×10472.27×1039\frac{F_e}{F_g} = \frac{2.304 \times 10^{-8}}{1.015 \times 10^{-47}} \approx 2.27 \times 10^{39}

Same answer. The symbolic cancellation method is cleaner in an exam, but this numerical route is a good sanity check.

In NCERT Class 12 Chapter 1 and JEE theory questions, they often ask you to “justify why gravity is neglected at atomic scale.” The answer is this ratio — 1039\sim 10^{39}. Memorise the order of magnitude, not the exact digits.


Common Mistake

Students forget to square the charge when computing ke2ke^2. They write k×ek \times e instead of k×e2k \times e^2. Both forces have the form constant×property2r2\text{constant} \times \frac{\text{property}^2}{r^2}, so the charge and mass each appear squared in their respective force formulas. Writing Fe=ke/r2F_e = ke/r^2 is dimensionally wrong and will cost you full marks in board exams.

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