Question

To simplify the calculation of a model with many interacting particles, after some threshold value ?=?,r=R, we approximate F as zero.

1. Explain the physical reasoning behind this assumption.
2. What is the force equation?
3. Evaluate the force F using both Coulomb’s law and our approximation, assuming two protons with a charge magnitude of 1.6022×10−19coulombs (C),1.6022×10−19coulombs (C), and the Coulomb constant ??=8.988×10^9Nm2/C2 ke=8.988×10^9Nm2/C2 are 1 m apart. Also, assume ?<1m. R<1m. How much inaccuracy does our approximation generate? Is our approximation reasonable?
4. Is there any finite value of R for which this system remains continuous at R?
5. Instead of making the force 0 at R, instead we let the force be 10⁻²⁰ for ?≥?.r≥R. Assume two protons, which have a magnitude of charge 1.6022×10^−19C,1.6022×10^−19C, and the Coulomb constant ??=8.988×10^9Nm2/C2. ke=8.988×10^9Nm2/C2. Is there a value R that can make this system continuous? If so, find it.

Answer 1

a) According to Coulomb’s law as the distance between the particles increases, then the force between them decreases. So, there must be a point at which the force between the particles is approximately 0 for larger values of distance.

b) The force between 2 charged particles is given by,

where r is the distance between them

q₁ is the charge of first particle

q₂ is the charge of second particle

k_e is the coulomb constant.

c) Here k_e=8.988 * 10⁹

q₁=q₂=1.6022*10^-19

r=1

Calculate force between them

The magnitude of the force is very less. So, it can be considered as 0.