Question

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. ${10}^{-47}C$
• B. ${10}^{-20}C$
• C. ${10}^{-23}C$
• D. ${10}^{-37}C$

Answer 1

The correct option is D ${10}^{-37}C$

Given : $m_h=1.67\times {10}^{-27}kg$
Net charge on hydrogen atom $q=(e+\Delta e)-e=\Delta e$
Let the distance between them be $d$.
Equating gravitational force and the electrostatic force, we get
$\frac{G{m}_h^2}{d^2}=\frac{k(\Delta e{)}^2}{d^2}$
Or $\frac{(6.67\times {10}^{-11})(1.67\times {10}^{-27}{)}^2}{d^2}=\frac{(9\times {10}^9)(\Delta e{)}^2}{d^2}$
$⟹\Delta e=2.06\times {10}^{-37}C$
The correct answer is option D.