Two tiny particles carrying like charges of the same magnitude are $1.0 \mathrm {~mm}$ apart. If the electric force on one of them is $5.0 \mathrm {~N}$, what is the magnitude of the charge on each of these particles? $( k =$ $1 / 4 \pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 }$ )

Question

Quizplus · Accepted Answer

The electric force between two charges is given by Coulomb's law: $F = k \frac{q_1 q_2}{r^2}$. Given that the charges are like and of the same magnitude, we can set $q_1 = q_2 = q$, thus $F = k \frac{q^2}{r^2}$. Rearranging for $q$, we get $q = \sqrt{\frac{F r^2}{k}}$. Substituting the given values: $q = \sqrt{\frac{5.0 \times (1.0 \times 10^{-3})^2}{9.0 \times 10^9}} = \sqrt{\frac{5.0 \times 10^{-6}}{9.0 \times 10^9}} = \sqrt{5.56 \times 10^{-16}} \approx 2.4 \times 10^{-8} \, \mathrm{C}$.

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 $1.0 \mathrm {~mm}$