Question

Suppose the charge of a proton and an electron differ slightly. One of them is $-e$ and the other is $(e+△e)$. If the net of electrostatic force and gravitational force between two hydrogen atoms places at a distance $d$ (much greater than atomic size) apart is zero, then $△e$ is of the order [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. ${10}^{-47}C$

Answer 1

Answer: (C)

${10}^{-37}C$

$\begin{array}{c}{q}_1=e\\ q_2=(e+\Delta e)\end{array}$

$\begin{array}{cc}{f}_{eq}& =\frac{1}{4\pi {\in }_0}2\{\frac{ex(e+\Delta e)}{d^2}-\frac{exe}{d^2}\}\\ & =18\times 10\left[\frac{e\Delta e}{d^2}\right]\\ F_{\text{net}}& =\frac{G\times m_1^2}{d^2}\end{array}$

$\begin{array}{c}{F}_{eq}=F_{net}\\ 18\times {\mathrm{10}}^9\left[\frac{e\Delta e}{d^2}\right]=\frac{6.67\times {10}^{-11}\times m_n^2}{d^2}\\ \Delta e=\frac{6.67\times {10}^{-11}\times {(1.67\times {10}^{-27})}^2}{18\times {10}^9\times 1.67\times {10}^{-19}}\\ ={10}^{-37}c\end{array}$