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 ascertain the value of \( P \), the electrostatic force is compared to the gravitational force between the ions. The formulas for these forces are:

1. Electrostatic force \( F_e \) between two charges: \(F_e = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\)
   • \( k = 8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) (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: \(F_g = \frac{G \cdot m_1 \cdot m_2}{r^2}\)
   • \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) (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 electrostatic force magnitude to the gravitational force magnitude is:

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

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 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 \( P \):

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

Therefore, the value of \( P \) is 10.