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 of the separation is increased to 25 cm?

Answer 1

Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other.
So, $$F_1=\frac{1}{4\pi {\in }_0}\cdot \frac{q^2}{r_1^2}$$

Also, $$F_2=\frac{1}{4\pi {\in }_0}\cdot \frac{q^2}{r_2^2}$$

According to the question, we have :

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

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

$$\therefore F_2=\frac{16}{25}\times F_1$$

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

$$\Rightarrow F_2=12.8N\approx 13.0N$$

Therefore, the two charged particles will exert a force of 13.0 N on each other, if the separation is increased to 25 cm.