what happens to the force acting between the charged particles, if the distance between these charged particles is halved?

what happens to the force acting between the charged particles, if the distance between these charged particles is halved?

1 hour ago 6
Nature

According to Coulomb's Law, the electrostatic force FFF between two charged particles with charges q1q_1q1​ and q2q_2q2​ separated by a distance rrr is given by:

F=k∣q1q2∣r2F=k\frac{|q_1q_2|}{r^2}F=kr2∣q1​q2​∣​

where kkk is Coulomb's constant. If the distance rrr between the two charged particles is halved, the new distance becomes r/2r/2r/2. Substituting this into the formula gives:

F′=k∣q1q2∣(r/2)2=k∣q1q2∣r2/4=4×k∣q1q2∣r2=4FF'=k\frac{|q_1q_2|}{(r/2)^2}=k\frac{|q_1q_2|}{r^2/4}=4\times k\frac{|q_1q_2|}{r^2}=4FF′=k(r/2)2∣q1​q2​∣​=kr2/4∣q1​q2​∣​=4×kr2∣q1​q2​∣​=4F

This means the force between the two charged particles becomes four times greater when the distance between them is halved

Summary:

  • Force is inversely proportional to the square of the distance.
  • Halving the distance increases the force by a factor of 444.
  • Thus, the force becomes four times the original force.

This inverse square relationship is a fundamental characteristic of Coulomb's law and is similar to the behavior of gravitational forces but with much stronger magnitude for charges

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