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how far apart are a proton and electron if they exert an attractive force of 3n on one another

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Edufirst

Answer:

     [tex]distance=8.76\times 10^{-5}angstrom[/tex]

Explanation:

The 3N force must be attributed to electrostatic attraction between both charges.

The formula is known as Coulomb's law:

    [tex]F_E=k\times \dfrac{q_1\cdot q_2}{d^2}[/tex]

Where:

  • [tex]F_E[/tex] is the elecrostatic force: 3 N
  • k is the constant  for Coulomb's law ≈ 9.00 × 10⁹ N . m² / C²
  • d is the distance that separate the centers of the charges
  • [tex]q_1\text{ and }q_2[/tex] are the values of the charges: ≈ ± 1.6 × 10⁻¹⁹ C

When you are not interested in the direction of the electrostatic force you just use the magnitudes of the charges.

Substitute in the formula and solve for d:

      [tex]3N=9.00\times 10^9N\cdotm^2/C^2\times\dfrac{(1.6\cdot 10^{-19}C)^2}{d^2}[/tex]

      [tex]d=8.76\times 10^{-15}m^2[/tex]

That can be converted to angstroms with the conversion factor:

  • 1m = 10¹⁰ angstrom

[tex]d= 8.76\times 10^{-15}m\times 10^{10}angstrom/m=8.76\times 10^{-5}angstrom[/tex]

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