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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 distanced (much greater than atomic size) apart is zero, then $\Delta e$ is of the order of $[$ Given: mass of hydrogen $m_h = 1.67 \times 10^{- 27}\,\, kg]$
$10^{-23}\,\, C$
$10^{-37 }\,\,C$
$10^{-47} \,\,C$
$10^{-20}\,\, C$
Solution
A hydrogen atom consists of an electron and a proton.
$\therefore$ Charge on one hydrogen atom
$=q_{e}+q_{p}=-e+(e+\Delta e)=\Delta e$
Since a hydrogen atom carry a net charge $\Delta e$
$\therefore \quad$ Electrostatic force,
$F_{e} \frac{1}{4 \pi \varepsilon_{o}} \frac{(\Delta e)^{2}}{d^{2}}………(i)$
will act between two hydrogen atoms.
The gravitational force between two hydrogen atoms is given as
$F_{g}=\frac{G m_{h} m_{h}}{d^{2}}………(ii)$
since, the net force on the system is zero, $F_{e}=F_{g}$ Using eqns. $(i)$ and $(ii)$, we get
${\frac{(\Delta e)^{2}}{4 \pi \varepsilon_{o} d^{2}}=\frac{G m_{h}^{2}}{d^{2}}}$
${(\Delta e)^{2}=4 \pi \varepsilon_{o} G m_{h}^{2}}$
${=6.67 \times 10^{-11} \times\left(1.67 \times 10^{-27}\right)^{2} /\left(9 \times 10^{9}\right)}$
${\Delta e=10^{-37} \,\mathrm{C}}$
