Question

If the distance between the two charged particles are doubled, then the force will :

• A. remain same
• B. be one-fourth
• C. be halved
• D. be doubled

Answer 1

Answer: (B)

From Coulombs law
$F=\frac{1}{4\pi {\epsilon }_0}=\frac{q_1{q}_2}{r^2}$
where $q_1,q_2$ are charges separated at a distance $r$.
Given, $F_1=F,r_1=r,r_2=2r$
$\therefore \frac{F_1}{F_2}=\frac{r_2^2}{r_1^2}=\frac{{(2r)}^2}{r^2}=4$
$\Rightarrow F_2=\frac{F_1}{4}$
Hence, force becomes one-fourth.