Questions: A micro-transporter moves from one side of an evacuated chamber to another. It is powered by equal-magnitude, opposite-sign charges Q and -Q on top of two station points, as shown in (Figure 1). The transporter's mass is 1.0 g and it is carrying a positive charge of 6.0 × 10^-10 C. If a 7.0 × 10^-3-N friction force opposes the transporter's motion, what must be the magnitude of Q to make the transporter move with an initial acceleration of 1.0 m / s^2 ? Express your answer with the appropriate units.

Solution Steps

The transporter experiences three forces:

1. The electrostatic force due to the charges $Q$ and $-Q$.
2. The friction force opposing its motion.
3. The net force causing the acceleration.

The net force $F_{net}$ is given by Newton's second law: \(F_{net} = ma\) where $m$ is the mass of the transporter and $a$ is the acceleration. Given $m = 1.0 \, g = 1.0 \times 10^{-3} \, kg$ and $a = 1.0 \, m/s^2$, we have: \(F_{net} = (1.0 \times 10^{-3} \, kg)(1.0 \, m/s^2) = 1.0 \times 10^{-3} \, N\)

The net force is the difference between the electrostatic force $F_e$ and the friction force $F_f$: \(F_{net} = F_e - F_f\) Given $F_f = 7.0 \times 10^{-3} \, N$, we can find the electrostatic force: \(F_e = F_{net} + F_f = 1.0 \times 10^{-3} \, N + 7.0 \times 10^{-3} \, N = 8.0 \times 10^{-3} \, N\)

The electrostatic force on the transporter is given by: \(F_e = qE\) where $q$ is the charge of the transporter and $E$ is the net electric field at its position. We can find the electric field: \(E = \frac{F_e}{q} = \frac{8.0 \times 10^{-3} \, N}{6.0 \times 10^{-10} \, C} = \frac{4}{3} \times 10^7 \, N/C\)

The net electric field due to the two charges $Q$ and $-Q$ at the location of the transporter (charge $q$) can be approximated as the field due to two point charges: \(E = \frac{kQ}{r_1^2} - \frac{k(-Q)}{r_2^2}\), where $r_1$ and $r_2$ are the distances from the charges $+Q$ and $-Q$ to the charge $q$ respectively, and $k = 8.99 \times 10^9 \, N m^2/C^2$ is Coulomb's constant. Here $r_1 = 6.0 \, mm = 6.0 \times 10^{-3} \, m$ and $r_2 = 36.0 \, mm + 6.0 \, mm = 42.0 \, mm = 42.0 \times 10^{-3} \, m$. So, \(E = kQ (\frac{1}{r_1^2} + \frac{1}{r_2^2}) \).

Therefore, \(Q = \frac{E}{k (\frac{1}{r_1^2} + \frac{1}{r_2^2})} = \frac{\frac{4}{3} \times 10^7}{8.99 \times 10^9 (\frac{1}{(6 \times 10^{-3})^2} + \frac{1}{(42 \times 10^{-3})^2})} \approx 1.667 \times 10^{-6} C\)

Final Answer