Electric Charges and Fields - Result Question 3
3. 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 $m_h=1.67 \times 10^{-27} kg$ ]
[2017]
(a) $10^{-23} C$
(b) $10^{-37} C$
(c) $10^{-47} C$
(d) $10^{-20} C$
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Answer:
Correct Answer: 3. (b)
Solution:
- (b) Net charge on one Hatom $=-e+(e+\Delta e)=\Delta e$ According to question, the net electrostatic force $(F_E)=$ gravitational force $(F_G)$ $F_E=F_G$
or $\frac{1}{4 \pi \varepsilon_0} \frac{\Delta e^{2}}{d^{2}}=\frac{Gm^{2}}{d^{2}}$
$\Rightarrow \Delta e=m \sqrt{\frac{G}{k}}(\frac{1}{4 \pi \varepsilon_0}=k=9 \times 10^{9})$
$=1.67 \times 10^{-27} \sqrt{\frac{6.67 \times 10^{-11}}{9 \times 10^{9}}}$
$\Delta e \approx 1.436 \times 10^{-37} C$
