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

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:

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