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NEET 2017 · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from NEET 2017

Q67NEET 2017Difficult

Electric Charge & Coulomb's Law

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

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Answer: (b)

  1. Each hydrogen atom has net charge Δe\Delta e, so the electric repulsion must equal the gravitational attraction.
  2. k(Δe)2d2=Gmh2d2⇒Δe=mhGk\dfrac{k(\Delta e)^2}{d^2} = \dfrac{Gm_h^2}{d^2} \Rightarrow \Delta e = m_h\sqrt{\dfrac{G}{k}}
  3. Δe=1.67×10−276.67×10−119×109≈1.67×10−27×8.6×10−11≈1.4×10−37 C\Delta e = 1.67 \times 10^{-27}\sqrt{\dfrac{6.67 \times 10^{-11}}{9 \times 10^9}} \approx 1.67 \times 10^{-27} \times 8.6 \times 10^{-11} \approx 1.4 \times 10^{-37}\text{ C}
  4. So the answer is (b) 10−3710^{-37} C.
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