Suppose the charge of a proton and an electron differ slightly. One is -e , the other is e + e . If the net electrostatic force and gravitational forc - Physics MCQ NEET (UG) 2017 2026 | ExamTest.live

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Suppose the charge of a proton and an electron differ slightly. One is - e, the other is e + Δ e . If the net electrostatic force and gravitational force between two hydrogen atoms at distance d is zero, then Δ e is of the order of: (Mass of hydrogen m_{h} = 1.67 \times 1 0^{- 27} kg)

Solution & Step-by-step Explanation

Net charge on one hydrogen atom q = (e + Δ e) - e = Δ e .Electrostatic force F_{e} = \frac{k ( Δ e ) ^{2}}{d ^{2}} .Gravitational force F_{g} = \frac{G m _{h}^{2}}{d ^{2}} .Given F_{e} = F_{g} : k (Δ e)^{2} = G m_{h}^{2} (Δ e)^{2} = \frac{6.67 \times 1 0 ^{- 11} \times ( 1.67 \times 1 0 ^{- 27} ) ^{2}}{9 \times 1 0 ^{9}} (Δ e)^{2} \approx \frac{6.67 \times 2.8 \times 1 0 ^{- 65}}{9 \times 1 0 ^{9}} \approx 2 \times 1 0^{- 74} Δ e \approx 2 \times 1 0^{- 74} \approx 1 0^{- 37} C

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Suppose the charge of a proton and an electron differ slightly. One is - e, the other is e + Δ e . If the net electrostatic force and gravitational force between two hydrogen atoms at distance d is zero, then Δ e is of the order of: (Mass of hydrogen m_{h} = 1.67 \times 1 0^{- 27} kg)

1 0^{- 23} C

1 0^{- 37} C

1 0^{- 47} C

1 0^{- 20} C

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