Question

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

Answer 1

Answer:

     $distance=8.76\times 10^{-5}angstrom$

Explanation:

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

The formula is known as Coulomb's law:

    $F_E=k\times \dfrac{q_1\cdot q_2}{d^2}$

Where:

• $F_E$ 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
• $q_1\text{ and }q_2$ 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:

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

      $d=8.76\times 10^{-15}m^2$

That can be converted to angstroms with the conversion factor:

• 1m = 10¹⁰ angstrom

$d= 8.76\times 10^{-15}m\times 10^{10}angstrom/m=8.76\times 10^{-5}angstrom$