When placed at a separation of 20 cm , two charged particles exert a Coulomb force of 20 N on each other. Determine the force if the separation is increased to 25 cm .

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Quantrex Academy · Accepted Answer

Correct answer: D. 12.8 N

According to Coulomb's law, the electrostatic force between two point charges is inversely proportional to the square of the distance between them, i.e.,

F \propto \dfrac{1}{r^2}

.

Therefore,

\dfrac{F_2}{F_1} = \left(\dfrac{r_1}{r_2}\right)^2

Given

F_1 = 20 \text{ N}

,

r_1 = 20 \text{ cm}

, and

r_2 = 25 \text{ cm}

.

Substituting the values:

F_2 = 20 \times \left(\dfrac{20}{25}\right)^2

F_2 = 20 \times \left(\dfrac{4}{5}\right)^2

F_2 = 20 \times \dfrac{16}{25} = \dfrac{64}{5} = 12.8 \text{ N}

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