Question

Two cylindrical glass beads each of mass \(m=10.0 \mathrm{mg}\) are set on their flat ends on a horizontal insulating surface separated by a distance \(d=2.00 \mathrm{~cm} .\) The coefficient of static friction between the beads and the surface is \(\mu_{s}=0.200\). The beads are then given identical charges (magnitude and sign). What is the minimum charge needed to start the beads moving?

Short answer

Answer: The minimum electric charge needed to start the motion of the two cylindrical glass beads is 1.24 × 10⁻⁹ C.

Step-by-step solution

Calculate the maximum static friction force

As the beads are placed on a horizontal insulating surface, the maximum static friction force (\(F_{max}\)) that resists their motion can be calculated using the formula \(F_{max} = \mu_s \cdot N\), where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force acting on the beads. Since the beads are on a horizontal surface, \(N = mg\), where \(m\) is the mass of the glass beads and \(g\) is the acceleration due to gravity (approximately \(9.81\,\mathrm{m/s^2}\)). Therefore, the maximum static friction force becomes: \(F_{max} = \mu_s \cdot mg\) Given, \(\mu_s = 0.200\) and \(m = 10.0\,\mathrm{mg} = 10.0 \times 10^{-6}\,\mathrm{kg}\), we can calculate: \(F_{max}= 0.200 \times (10.0 \times 10^{-6} \,\mathrm{kg}) \times 9.81 \,\mathrm{m/s^2} = 1.962 \times 10^{-5}\,\mathrm{N}\).

Calculate the electric force between the beads

To determine the electric force between the beads, we will use Coulomb's law. Coulomb's law states that the magnitude of the force (\(F_e\)) between two particles with charges \(q_1\) and \(q_2\) separated by a distance \(r\) can be calculated as: \(F_e = k \cdot \frac{|q_1q_2|}{r^2}\) Where \(k\) is Coulomb's constant and equals \(8.99 \times 10^9\,\mathrm{N\,m^2/C^2}\), and \(r\) is the distance between the charges. In our case, \(q_1 = q_2 = q\) (since the beads have identical charges) and \(r = d = 2.00\,\mathrm{cm} = 2.00\times 10^{-2}\,\mathrm{m}\). Substituting these values into Coulomb's law, we get: \(F_e = k \cdot \frac{q^2}{d^2} = \frac{8.99 \times 10^9\,\mathrm{N\,m^2/C^2} \cdot q^2}{(2.00\times 10^{-2}\,\mathrm{m})^2}\)

Equate the electric force to the maximum static friction force

As we need to find the minimum charge that will make the beads start moving, we can equate the electric force to the maximum static friction force: \(F_e = F_{max}\) Substituting the expressions for \(F_e\) and \(F_{max}\), we have: \(\frac{8.99 \times 10^9\,\mathrm{N\,m^2/C^2} \cdot q^2}{(2.00\times 10^{-2}\,\mathrm{m})^2} = 1.962 \times 10^{-5}\,\mathrm{N}\)

Solve for the minimum charge

Now, we can solve the equation for the minimum charge \(q\): \(q^2 = \frac{1.962 \times 10^{-5}\,\mathrm{N} \cdot (2.00\times 10^{-2}\,\mathrm{m})^2}{8.99 \times 10^9\,\mathrm{N\,m^2/C^2}}\) Calculating this expression and taking the square root of both sides, we find: \(q = 1.24 \times 10^{-9}\,\mathrm{C}\) So, the minimum charge needed to start the beads moving is \(1.24 \times 10^{-9}\,\mathrm{C}\).