JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A positive ion \(A\) and a negative ion \(B\) has charges \(6.67 \times 10^{-19} \mathrm{C}\) and \(9.6 \times 10^{-10} \mathrm{C}\), and masses \(19.2 \times 10^{-27} \mathrm{~kg}\) and \(9 \times 10^{-27} \mathrm{~kg}\) 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 \(P \times 10^{45}\), where the value of 10\(P\) is _______.
(Take \(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-1}\) and universal gravitational constant as \(6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\) )
Assume that charge may not be an integral multiple of electrons.
- A 5
- B 10
- C 15
- D 20
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation
\( \frac{F_e}{F_g} = \frac{k |q_A q_B|}{G m_A m_B} \) \( \frac{F_e}{F_g} = \frac{(9 \times 10^9)(6.67 \times 10^{-19})(9.6 \times 10^{-10})}{(6.67 \times 10^{-11})(19.2 \times 10^{-27})(9 \times 10^{-27})} \)…
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