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

Answer 1

Answer: (B)

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{ Gm _{ h }^2}{ d ^2}=\frac{ k (\Delta e )^2}{ d ^2} $
Or $\frac{\left(6.67 \times 10^{-11}\right)\left(1.67 \times 10^{-27}\right)^2}{ d ^2}=\frac{\left(9 \times 10^9\right)(\Delta e )^2}{ d ^2}$
$\Longrightarrow \Delta e =2.06 \times 10^{-37} C$