What is the repulsive electrical force between two protons 4.0×10−15m4.0×10−15m apart from each other in an atomic nucleus?
Question
What is the repulsive electrical force between two protons 4.0×10−15m4.0×10−15m apart from each other in an atomic nucleus?
Solution
The force between two charges is given by Coulomb's Law:
F = k * q1 * q2 / r^2
where:
- F is the force between the charges,
- k is Coulomb's constant (8.99 * 10^9 N m^2/C^2),
- q1 and q2 are the amounts of charge (for a proton, this is 1.6 * 10^-19 C), and
- r is the separation between the charges.
Substituting the given values:
F = (8.99 * 10^9 N m^2/C^2) * (1.6 * 10^-19 C)^2 / (4.0 * 10^-15 m)^2
Solving this equation will give you the force in Newtons.
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