Q. Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e + $\Delta$e). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then $\Delta$e is of the order of [Given mass of hydrogen $m_h = 1.67 \times 10^{-27}$ kg]
A) $10^{-20}$ C
B) $10^{-23}$ C
C) $10^{-37}$ C
D) $10^{-47}$ C
Explanation
[Solution] $\frac{(\Delta e)^2}{4\pi\epsilon_0 d^2} = \frac{Gm_h^2}{d^2}$, so $\Delta e = m_h\sqrt{4\pi\epsilon_0 G} \approx 1.67 \times 10^{-27} \times 8.6 \times 10^{-11} \approx 10^{-37}$ C. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.