Question

Two 1.0 g beads are charged equally and placed 5.0 cm apart. When released, they begin to accelerate at 150m/s^2. What is the magnitude of the charge on each bead?

Answer 1

Answer on Question #43713 – Physics – Mechanics | Kinematics | Dynamics

1. Two $1.0\,\mathrm{g}$ beads are charged equally and placed $5.0\,\mathrm{cm}$ apart. When released, they begin to accelerate at $150\,\mathrm{m/s^2}$. What is the magnitude of the charge on each bead?

This force is $F = -G \frac{m^2}{d^2} + k \frac{q^2}{d^2}$.

So, we can write down: $ma = -G \frac{m^2}{d^2} + k \frac{q^2}{d^2}$.

The charge on each bead is $\sqrt{q = d \sqrt{\left(a + G \frac{m}{d^2}\right) \frac{m}{k}}}$.

Let check the dimension: $[q] = m \sqrt{\left(\frac{m}{s^2} + \frac{N \cdot m^2}{kg^2} \cdot \frac{kg}{m^2}\right) \frac{kg}{N \cdot m^2 / C^2}} = C$.

Let evaluate the quantity: $q = 0.05 \cdot \sqrt{\left(150 + 6.67 \cdot 10^{-11} \cdot \frac{0.001}{0.05^2}\right) \cdot \frac{0.001}{9 \cdot 10^9}} = 2.04 \cdot 10^{-4}(C)$.

Answer: $204\,\mathrm{mC}$.

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