JEE Main 2025 Jan 23 Shift 1, Physics Q21: Coulomb's Law And Electric Field
A positive ion A and a negative ion B have charges \(6.67 \times 10^{-19}\,\text{C}\) and \(11.52 \times 10^{-19}\,\text{C}\), and masses \(19.2 \times 10^{-27}\,\text{kg}\) and \(9 \times 10^{-27}\,\text{kg}\) 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 \(P \times 10^{35}\), where the value of \(P\) is ____. (Take \(\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\text{N m}^{2}\text{C}^{-2}\) and universal gravitational constant as \(6.67\times10^{-11}\,\text{N m}^{2}\text{kg}^{-2}\).)
Given:
\(q_1=6.67\times10^{-19}\,\text{C}\)
\(q_2=11.52\times10^{-19}\,\text{C}\)
\(m_1=19.2\times10^{-27}\,\text{kg}\)
\(m_2=9\times10^{-27}\,\text{kg}\)
\(k=9\times10^{9}\,\text{N m}^2\text{C}^{-2}\)
\(G=6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}\)
The ratio of electrostatic force to gravitational force is
\[ \dfrac{F_e}{F_g} = \dfrac{\dfrac{kq_1q_2}{r^2}} {\dfrac{Gm_1m_2}{r^2}} = \dfrac{kq_1q_2}{Gm_1m_2}. \]
Substituting the given values,
\[ \dfrac{F_e}{F_g} = \dfrac{(9\times10^{9})(6.67\times10^{-19})(11.52\times10^{-19})} {(6.67\times10^{-11})(19.2\times10^{-27})(9\times10^{-27})}. \]
Cancel the common factors \(6.67\) and \(9\):
\[ \dfrac{F_e}{F_g} = \dfrac{11.52\times10^{-29}} {19.2\times10^{-65}}. \]
Simplifying,
\[ \dfrac{11.52}{19.2}=0.6 \]
\[ 10^{-29}\div10^{-65}=10^{36} \]
Therefore,
\[ \dfrac{F_e}{F_g} = 0.6\times10^{36} = 6\times10^{35}. \]
Comparing with \(P\times10^{35}\), we get
\[ \boxed{P=6} \]
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