Question:medium

For any fixed distance, the electromagnetic force between two protons is \( 10^n \) times of the gravitational force between them. Then \( n = \)

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Remember these standard ratios: For electron-electron, ratio is \( \approx 10^{42} \); for proton-proton, ratio is \( \approx 10^{36} \).
Updated On: Mar 30, 2026
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Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
We need to compare the electrostatic force and gravitational force between two protons separated by a distance \( r \). We use Coulomb's Law and Newton's Law of Gravitation.
Step 2: Key Formula or Approach:
1. Electrostatic Force: \( F_e = \frac{k e^2}{r^2} \), where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \). 2. Gravitational Force: \( F_g = \frac{G m_p^2}{r^2} \), where \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \). 3. Ratio: \( \frac{F_e}{F_g} = \frac{k e^2}{G m_p^2} \).
Step 3: Detailed Explanation:
Given values: - Charge of proton \( e = 1.6 \times 10^{-19} \, \text{C} \) - Mass of proton \( m_p = 1.67 \times 10^{-27} \, \text{kg} \) Calculate the ratio: \[ \frac{F_e}{F_g} = \frac{(9 \times 10^9) (1.6 \times 10^{-19})^2}{(6.67 \times 10^{-11}) (1.67 \times 10^{-27})^2} \] \[ = \frac{9 \times 2.56 \times 10^9 \times 10^{-38}}{6.67 \times 2.79 \times 10^{-11} \times 10^{-54}} \] \[ = \frac{23.04 \times 10^{-29}}{18.6 \times 10^{-65}} \] \[ \approx 1.24 \times 10^{-29 - (-65)} \] \[ \approx 1.24 \times 10^{36} \] The order of magnitude is \( 10^{36} \). So, \( n \approx 36 \).
Step 4: Final Answer:
The value of \( n \) is 36.
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