Answer :
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]