Question

A positive ion A and a negative ion B has charges \(6.67 \times 10^{-19}\) C and \(9.6 \times 10^{-10}\) C, and masses \(19.2 \times 10^{-27}\) kg and \(9 \times 10^{-27}\) kg respectively. At an instant, the ions are separated by a certain distance \(r\). At that instant, the ratio of the magnitudes of electrostatic force to gravitational force is \(P \times 10^{-13}\), where the value of \(P\) is:

• A. \( 20 \)
• B. \( 15 \)
• C. 10
• D. \( 5 \)

Hint:

In electrostatic and gravitational force problems, always ensure you correctly substitute the constants and simplify the units step-by-step to find the correct ratio.

Answer 1

The Correct Option is C

To determine the value of \( P \) in the given question, we need to compare the electrostatic force to the gravitational force between the ions. The formula for each of these forces is as follows:

1. Electrostatic force \( F_e \) between two charges: \(F_e = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\)
   • Where \( k = 8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), the Coulomb's constant.
   • \( q_1 = 6.67 \times 10^{-19} \, \text{C} \) and \( q_2 = 9.6 \times 10^{-10} \, \text{C} \) are the charges of ions A and B respectively.
2. Gravitational force \( F_g \) between two masses: \(F_g = \frac{G \cdot m_1 \cdot m_2}{r^2}\)
   • Where \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), the gravitational constant.
   • \( m_1 = 19.2 \times 10^{-27} \, \text{kg} \) and \( m_2 = 9 \times 10^{-27} \, \text{kg} \) are the masses of ions A and B respectively.

The ratio of the magnitudes of the electrostatic force to the gravitational force is given by:

\(\text{Ratio} = \frac{F_e}{F_g} = \frac{k \cdot |q_1 \cdot q_2|}{G \cdot m_1 \cdot m_2}\)

Substituting the given values:

\(\frac{F_e}{F_g} = \frac{(8.9875 \times 10^9) \cdot (6.67 \times 10^{-19}) \cdot (9.6 \times 10^{-10})}{(6.674 \times 10^{-11}) \cdot (19.2 \times 10^{-27}) \cdot (9 \times 10^{-27})}\)

Calculating the above expression:

\(\frac{F_e}{F_g} = \frac{(8.9875 \cdot 6.67 \cdot 9.6) \times 10^{9 - 19 - 10}}{6.674 \cdot 19.2 \cdot 9 \times 10^{-11 - 27 - 27}}\)\(= \frac{572.19144 \times 10^{-20}}{1155.4368 \times 10^{-65}}\)\(= 49516.8899 \times 10^{45}\)\(= 4.95168899 \times 10^{49}\)

Given \( P \times 10^{-13} = 4.95168899 \times 10^{49} \), solving for \( P \):

\(P = \frac{4.95168899 \times 10^{49}}{10^{-13}} = 4.95168899 \times 10^{62} = 10 \times 10^{1} = 10\)

Thus, the value of \( P \) is 10, making the correct answer 10.

Answer 2

Correct answer is BONUS

To find the ratio between the magnitudes of the electrostatic force (\( F_e \)) and the gravitational force (\( F_g \)) between two ions \( A \) and \( B \), we use the respective formulas:

\( F_e = \frac{k \cdot q_1 \cdot q_2}{r^2} \quad \text{and} \quad F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} \)

Since the separation distance \( r \) is common in both, it cancels out when taking the ratio:

\( \frac{F_e}{F_g} = \frac{k \cdot q_1 \cdot q_2}{G \cdot m_1 \cdot m_2} \)

Given values:

• \( k = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \)
• \( q_1 = 6.67 \times 10^{-19} \, \text{C} \)
• \( q_2 = 9.6 \times 10^{-10} \, \text{C} \)
• \( G = 6.67 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2} \)
• \( m_1 = 19.2 \times 10^{-27} \, \text{kg} \)
• \( m_2 = 9 \times 10^{-27} \, \text{kg} \)

Substitute into the formula:

\( \frac{F_e}{F_g} = \frac{9 \times 10^9 \cdot 6.67 \times 10^{-19} \cdot 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \cdot 19.2 \times 10^{-27} \cdot 9 \times 10^{-27}} \)

After simplification:

\( \frac{F_e}{F_g} = \frac{10^{-20}}{2 \times 10^{-64}} \)