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

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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