Two tiny particles carrying like charges of the same magnitude are $1.0 \mathrm {~mm}$ apart.If the electric force exerted on one of them is $5.0 \mathrm {~N} \text {, }$ what is the magnitude of the charge on each of these particles?(k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

The electric force between two charges can be calculated using Coulomb's law: $F = k \frac{q_1 q_2}{r^2}$. Given that the charges are like and have the same magnitude, we can set $q_1 = q_2 = q$, and rearrange the formula to solve for $q$: $q = \sqrt{\frac{F r^2}{k}}$. Substituting the given values ($F = 5.0 \, \text{N}$, $r = 1.0 \times 10^{-3} \, \text{m}$, $k = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$), we find $q = 2.4 \times 10^{-8} \, \text{C}$.

Two Tiny Particles Carrying Like Charges of the Same Magnitude 1.0 mm1.0 \mathrm {~mm}1.0 mm

Two Tiny Particles Carrying Like Charges of the Same Magnitude $1.0 \mathrm {~mm}$