Question

Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other. What will be the force if the separation is increased to 25 cm ?

Answer 1

Here $F_1=\frac{1}{4\pi {\in }_0}.\frac{q^2}{r_1^2}$

and $F_2=\frac{1}{4\pi {\in }_0}.\frac{q^2}{r_2^2}$

$\therefore \frac{F_2}{F_1}=\frac{r_1^2}{r_2^2}$

$=\frac{20\times 20}{25\times 25}=\frac{16}{25}$

$\therefore F_2=\frac{16}{25}\times 20$

$=12.8N$