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

The Correct Option is B

$F_{e}=F_{g}$
$\frac{1}{4 \pi \varepsilon_{0}} \frac{\Delta e^{2}}{d^{2}}=\frac{G m^{2}}{d^{2}}$
$9 \times 10^{9}\left(\Delta e^{2}\right)=6.67 \times 10^{-11} \times 1.67$
$\times 10^{-27} \times 1.67 \times 10^{-27}$
$\Delta e^{2}=\frac{6.67 \times 1.67 \times 1.67}{9} \times 10^{-74}$
$\Delta e \approx 10^{-37}$