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?

#### Solution

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 :

\[\frac{F_2}{F_1} = \frac{r_1^2}{r_2^2}\]

\[ = \frac{20 \times 20}{25 \times 25} = \frac{16}{25}\]

\[ \therefore F_2 = \frac{16}{25} \times F_1\]

\[\Rightarrow F_2 = \frac{16}{25} \times 20 \]

\[ \Rightarrow F_2 = 12 . 8 N \approx 13 . 0 N\]

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