Suppose the charge of a proton and an electron differ slightly. One of them is e , the other is e + Δ e . If the net electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δ e is of the order of [Given mass…

Question

Suppose the charge of a proton and an electron differ slightly. One of them is e , the other is e + Δ e . If the net electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δ e is of the order of [Given mass of hydrogen m h = 1.67 × 10 – 27 kg ]

Aicharya · Accepted Answer

The net electrostatic force and gravitational force will be equal since the net force on the system is zero. Thus,

F_{E} = F_{g}

\frac{1}{4 π ε_{0}} \frac{Δ e^{2}}{d^{2}} = \frac{G m^{2}}{d^{2}}

9 \times 10^{9} (Δ e^{2}) = 6.67 \times 10^{- 11} \times 1.67 \times 10^{- 27} \times 1.67 \times 10^{- 27}

Δ e^{2} = \frac{6.67 \times 1.67 \times 1.67}{9} \times 10^{- 74}

Δ e \approx 10^{- 37}

Suppose the charge of a proton and an electron differ slightly. One of them is e , the other is e + Δ e . If…

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