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Physics Question on Electric charges and fields

Suppose the charge of a proton and an electron differ slightly. One of them is −e- e, the other is (e+Δee + \Delta e ). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance dd (much greater than atomic size) apart is zero, then Δe\Delta e is of the order of [Given mass of hydrogen mh=1.67×10−27 kgm_h = 1.67 \times 10^{-27} \, kg]

A

10−23C10^{-23} C

B

10−37C10^{-37} C

C

10−47C10^{-47} C

D

10−20C10^{-20} C

Answer

10−37C10^{-37} C

Explanation

Solution

Fe=FgF_{e}=F_{g}
14πε0Δe2d2=Gm2d2\frac{1}{4 \pi \varepsilon_{0}} \frac{\Delta e^{2}}{d^{2}}=\frac{G m^{2}}{d^{2}}
9×109(Δe2)=6.67×10−11×1.679 \times 10^{9}\left(\Delta e^{2}\right)=6.67 \times 10^{-11} \times 1.67
×10−27×1.67×10−27\times 10^{-27} \times 1.67 \times 10^{-27}
Δe2=6.67×1.67×1.679×10−74\Delta e^{2}=\frac{6.67 \times 1.67 \times 1.67}{9} \times 10^{-74}
Δe≈10−37\Delta e \approx 10^{-37}