When the distance between two charged particles is halved, the electrostatic force between them becomes four times larger.

Quick Scoop

Using Coulomb’s law, the force between two point charges is

F=kq1q2r2F=k\frac{q_1q_2}{r^2}F=kr2q1​q2​​

where rrr is the distance between them.

If the distance is halved, the new distance is r′=r2r'=\frac{r}{2}r′=2r​.

F′=kq1q2(r′)2=kq1q2(r2)2=kq1q2r24=4 kq1q2r2=4FF'=k\frac{q_1q_2}{(r')^2} =k\frac{q_1q_2}{\left(\frac{r}{2}\right)^2} =k\frac{q_1q_2}{\frac{r^2}{4}} =4,k\frac{q_1q_2}{r^2} =4FF′=k(r′)2q1​q2​​=k(2r​)2q1​q2​​=k4r2​q1​q2​​=4kr2q1​q2​​=4F

So the new force F′F'F′ is four times the original force FFF.

One-line answer for exams

When the distance between the charged particles is halved, the force between them becomes four times the original value (increases by a factor of 4).

Tiny example to remember

  • Imagine the force was 222 N at distance rrr.
  • Halve the distance to r/2r/2r/2.
  • The new force becomes 4×2=84\times 2=84×2=8 N.

A handy memory trick:

Distance ×2 ⇒ force ÷4
Distance ÷2 ⇒ force ×4

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Find out what happens to the force acting between charged particles if the distance between them is halved, using Coulomb’s law and a simple step-by-step explanation.

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