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

The magnitude of the electric force between two protons at this distance is mathematically given as

F=61.8N

What is the magnitude of the electric force (in N) between two protons at this distance?

Question Parameter(s):

The protons in a nucleus are approximately 2 ✕ 10^−15 m apart.

a distance d = 1.93 ✕ 10^−15 m apart.

Generally, the equation for the electric force (F)   is mathematically given as

$F=k.\frac{q_{1}q_{2}}{d^{2} }$

Therefore

$F=8.99* 10^{9} N.m^{2} .C^{-2} .\frac{(1.60 * 10^{-19}C)^{2} }{(1.93* 10^{-15}m)^{2} }$

F=61.8N

In conclusion, The force is

F=61.8N

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