Question

The protons in a nucleus are approximately 2 ✕ 10^−15 m apart. Consider the case where the protons are a distance d = 1.93 ✕ 10^−15 m apart. Calculate the magnitude of the electric force (in N) between two protons at this distance.

Answer 1

Answer:

61.8 N

Explanation:

Given data

• Charge of the protons (q): 1.60 × 10⁻¹⁹ Coulomb

• Distance between the protons (d): 1.93 × 10⁻¹⁵ meters

• Coulomb's constant (k): 8.99 × 10⁹ N.m².C⁻²

We can find the magnitude of the electric force (F) between the two protons using Coulomb's law.

$F=k.\frac{q_{1}q_{2}}{d^{2} } \\F=8.99 \times 10^{9} N.m^{2} .C^{-2} .\frac{(1.60 \times 10^{-19}C)^{2} }{(1.93 \times 10^{-15}m)^{2} } \\F=61.8N$

The magnitude of the electric force is 61.8 N.

Answer 2

Answer: 61.2N

F=kq1q2/r^2

F=8.98E9(1.6E-19)(1.6E-19)/(1.93E-15)^2

F=61.2N