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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 23 January · Morning Shift · Physics · Electrostatics

A positive ion A and a negative ion B has charges 6.67 × 10^ -19 C...

A positive ion AA and a negative ion BB has charges 6.67×10−19C6.67 \times 10^{-19} \mathrm{C} and 9.6×10−10C9.6 \times 10^{-10} \mathrm{C}, and masses 19.2×10−27 kg19.2 \times 10^{-27} \mathrm{~kg} and 9×10−27 kg9 \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×1045P \times 10^{45}, where the value of 10PP is \qquad (Take 14πε0=9×109Nm2C−1\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-1} and universal gravitational constant as 6.67×10−11Nm2 kg−26.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2} ) Assume that charge may not be an integral multiple of electrons.

Correct answer

0

Step-by-step solution

Fe=kq1q2r2Fg=Gm1m2r2FeFg=kq1q2Gm1m2=9×109×6.67×10−19×9.6×10−106.67×10−11×19.2×10−27×9×10−27=10−202×10−65\begin{aligned} F_e & =\frac{k q_1 q_2}{r^2} \\ F_g & =\frac{G m_1 m_2}{r^2} \\ \frac{F_e}{F_g} & =\frac{k q_1 q_2}{G m_1 m_2} \\ & =\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}} \\ & =\frac{10^{-20}}{2 \times 10^{-65}} \end{aligned}
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