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?

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Solution

Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other.
So, $F_{1} = \frac{1}{4 π \in_{0}} \cdot \frac{q^{2}}{r_{1}^{2}}$
Also, $F_{2} = \frac{1}{4 π \in_{0}} \cdot \frac{q^{2}}{r_{2}^{2}}$
According to the question, we have:
$\frac{F_{2}}{F_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{20 \times 20}{25 \times 25} = \frac{16}{25} ∴ F_{2} = \frac{16}{25} \times F_{1}$
$\Rightarrow F_{2} = \frac{16}{25} \times 20 \Rightarrow F_{2} = 12.8 N \approx 13.0 N$

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

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