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

We are given the following values: - \( q_1 = 6.67 \times 10^{-19} \, \text{C} \) (charge of ion A) - \( q_2 = 9.6 \times 10^{-10} \, \text{C} \) (charge of ion B) - \( m_1 = 19.2 \times 10^{-27} \, \text{kg} \) (mass of ion A) - \( m_2 = 9 \times 10^{-27} \, \text{kg} \) (mass of ion B) The electrostatic force \( F_{\text{ele}} \) between the two ions is given by Coulomb's law: \[ F_{\text{ele}} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \] where \( \epsilon_0 = 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \). The gravitational force \( F_{\text{grav}} \) between the two ions is given by Newton's law of gravitation: \[ F_{\text{grav}} = \frac{G m_1 m_2}{r^2} \] where \( G = 6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2} \). The ratio of the electrostatic force to the gravitational force is: \[ \frac{F_{\text{ele}}}{F_{\text{grav}}} = \frac{\frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}}{\frac{G m_1 m_2}{r^2}} = \frac{1}{4\pi \epsilon_0} \times \frac{q_1 q_2}{G m_1 m_2} \] Substituting the known values: \[ \frac{F_{\text{ele}}}{F_{\text{grav}}} = \frac{9 \times 10^9 \times (6.67 \times 10^{-19} \times 9.6 \times 10^{-10})}{6.67 \times 10^{-11} \times (19.2 \times 10^{-27} \times 9 \times 10^{-27})} \] Simplifying: \[ \frac{F_{\text{ele}}}{F_{\text{grav}}} = \frac{9 \times 10^9 \times 6.39 \times 10^{-28}}{6.67 \times 10^{-11} \times 1.728 \times 10^{-53}} \] \[ = \frac{5.751 \times 10^{-18}}{1.15 \times 10^{-64}} = 10^{45} \] Thus, \( P = 10 \). Thus, the value of \( P \) is \( \boxed{10} \).