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 4.0×10−15m 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.
Similar Questions
What is the electric potential energy of a system that consists of two protons 2.4×10−15 mm apart-as might occur inside a typical nucleus?
beam of electrons moves in astraight line parallel to the beam ofprotons in the same direction, as inthe figure. The force between themis….
An electron and a proton are placed 10 mm apart, the dipole moment of the system of charges is
What is stronger, the strong nuclear force or electromagnetic (electric repulsion) force?*1 pointStrong nuclearElectromagnetic
The ratio of charge and mass would be greatest forA Proton B Electron C Neutron D Alpha
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.