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Electrostatics question

2017 · Q124
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Electrostatics question

2017 · Q124

NEETPhysicsElectrostaticsMCQ+4 / −1
Suppose the charge of a proton and an electron differ slightly. One of them is −-−e, the other is (e + Δ\DeltaΔe). 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\DeltaΔe is of the order of
[Given : mass of hydrogen mh = 1.67 ×\times× 10−-−27 kg]
  1. A
    10−-−23 C
  2. B
    10−-−37 C
  3. C
    10−-−47 C
  4. D
    10−-−20 C
View written solutionFree

Correct answer: B

According to question, the net electrostatic force (FE) = gravitational force (FG)

FE = FG

⇒14πε0Δe2d2=Gm2d2 \Rightarrow {1 \over {4\pi {\varepsilon _0}}}{{\Delta {e^2}} \over {{d^2}}} = {{G{m^2}} \over {{d^2}}}⇒4πε0​1​d2Δe2​=d2Gm2​

⇒Δe=mGK(14πε0=k=9×109) \Rightarrow \Delta e = m\sqrt {{G \over K}} \left( {{1 \over {4\pi {\varepsilon _0}}} = k = 9 \times {{10}^9}} \right)⇒Δe=mKG​​(4πε0​1​=k=9×109)

=1.67×10−276.67×10−119×109= 1.67 \times {10^{ - 27}}\sqrt {{{6.67 \times {{10}^{ - 11}}} \over {9 \times {{10}^9}}}}=1.67×10−279×1096.67×10−11​​

Δe≈1.436×10−37C\Delta e \approx 1.436 \times {10^{ - 37}}CΔe≈1.436×10−37C

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