Question

The ratio of forces between two point charges in the two cases when their separation is halved and when it is doubled, is

• A. 1:02
• B. 4:01
• C. 8:01
• D. 16:01

Answer 1

Answer: (D)

16:01

Answer 2

According to Coulomb's law
$F = \frac{1 }{4 \pi \varepsilon_0 } \frac{q_1 q_2}{r^2}$
For same medium and charges, $F \propto \frac{1}{r^2}$
So, $\frac{F_{1}}{F_{2}} = \frac{r^{2}_{2}}{r_{1}^{2}}$
Here, $r_{1} = \frac{r}{2}$ and $r_{2} = 2r$
$\therefore \, \frac{F_{1}}{F_{2}} = \frac{\left(2r\right)^{2}}{\left(\frac{r}{2}\right)^{2}} = 16$