Two tiny beads, each of mass $3.2 \mathrm {~g}$, carry equal-magnitude charges. When they are placed $6.4 \mathrm {~cm}$ apart and released in outer space, they begin to accelerate toward each other at $538 \mathrm {~m} / \mathrm { s } ^ { 2 }$. What is the magnitude of the charge on each bead? $\left( k = 1 / 4 \pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right.$ )

Question

Quizplus · Accepted Answer

The force between the two charges can be calculated using Coulomb's law, $F = k \frac{q^2}{r^2}$, where $k = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, $q$ is the charge on each bead, and $r = 6.4 \, \text{cm} = 0.064 \, \text{m}$ is the distance between them. The force is also equal to the mass times the acceleration of each bead, $F = ma$, where $m = 3.2 \, \text{g} = 0.0032 \, \text{kg}$ and $a = 538 \, \text{m/s}^2$. Equating the two expressions for force gives $k \frac{q^2}{r^2} = ma$. Solving for $q$ gives $q = \sqrt{\frac{ma r^2}{k}}$. Substituting the given values, $q = \sqrt{\frac{0.0032 \cdot 538 \cdot (0.064)^2}{9.0 \times 10^9}} \approx 8.9 \times 10^{-7} \, \text{C} = 890 \, \text{nC}$.

Two Tiny Beads, Each of Mass 3.2 g3.2 \mathrm {~g}3.2 g , Carry Equal-Magnitude Charges

Two Tiny Beads, Each of Mass $3.2 \mathrm {~g}$ , Carry Equal-Magnitude Charges