The repulsive force between teo electrons has a magnitude of 16.0 N. What is the distance between the electrons
The correct answer and explanation is:
To find the distance between two electrons given the repulsive force, use Coulomb’s Law: F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2}
Where:
- FF = force between charges (16.0 N)
- kk = Coulomb’s constant 8.99×109 N\cdotpm2/C28.99 \times 10^9 \, \text{N·m}^2/\text{C}^2
- q1q_1 and q2q_2 = charge of electrons 1.60×10−19 C1.60 \times 10^{-19} \, \text{C}
- rr = distance between the electrons (unknown)
Rearranging the formula to solve for rr: r=k∣q1q2∣Fr = \sqrt{ k \frac{|q_1 q_2|}{F} }
Plugging in the values: r=8.99×109×(1.60×10−19)216.0r = \sqrt{ \frac{8.99 \times 10^9 \times (1.60 \times 10^{-19})^2}{16.0} }
First calculate the numerator inside the square root: (1.60×10−19)2=2.56×10−38(1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38}
Multiply by Coulomb’s constant: 8.99×109×2.56×10−38=2.299×10−288.99 \times 10^9 \times 2.56 \times 10^{-38} = 2.299 \times 10^{-28}
Now divide by the force: 2.299×10−2816.0=1.4369×10−29\frac{2.299 \times 10^{-28}}{16.0} = 1.4369 \times 10^{-29}
Finally, take the square root: r=1.4369×10−29=1.1987×10−15 metersr = \sqrt{1.4369 \times 10^{-29}} = 1.1987 \times 10^{-15} \text{ meters}
Rounded appropriately: r≈1.20×10−15 mr \approx 1.20 \times 10^{-15} \, \text{m}
Explanation:
The repulsive force between two electrons is due to their electric charges, which are identical and negative. The strength of this force can be calculated using Coulomb’s law, which relates the magnitude of the electric force to the product of the charges and inversely to the square of the distance between them. The constant kk accounts for the permittivity of free space and units.
Given the force and the charges, the only unknown is the distance rr. By isolating rr, the formula shows that the distance is proportional to the square root of the ratio of the product of the charges and the constant over the force.
Because the electron charge is extremely small, the forces become significant only when the electrons are very close to each other. The calculated distance 1.20×10−151.20 \times 10^{-15} meters is extremely small, on the order of a femtometer, which is comparable to the size of atomic nuclei, highlighting the strong repulsive force at very short ranges.
This calculation assumes the electrons are point charges in a vacuum and does not account for quantum effects, which become significant at such small scales. However, the classical Coulomb’s law gives a good approximation for the distance in this context.