Electrostatics

JEE Main level, numerical answer, medium

11 of 311

Question 11 of 311

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.

Show answer and solution

Answer: 5

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}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Electric field lines simulationDrag the sliders and watch the numbers in this question change.Revise Electrostatics firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Electrostatics (Electric Charge and Coulomb's law), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.