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Chapter 21: Problem 78

A bead with charge \(q_{1}=1.27 \mu \mathrm{C}\) is fixed in place at the end of a wire that makes an angle of \(\theta=51.3^{\circ}\) with the horizontal. A second bead with mass \(m_{2}=3.77 \mathrm{~g}\) and a charge of \(6.79 \mu \mathrm{C}\) slides without friction on the wire. What is the distance \(d\) at which the force of the Earth's gravity on \(m_{2}\) is balanced by the electrostatic force between the two beads? Neglect the gravitational interaction between the two beads.

Short Answer

Expert verified
Answer: The distance between the two charged beads when the gravitational force is balanced by the electrostatic force is 0.0206 m or 2.06 cm.

Step by step solution

01

Calculate the electrostatic force using Coulomb's law

Coulomb's law gives the force exerted between two charges as \( F_{e} = k \frac{q_1 q_2}{r^2} \), where: - \(F_{e}\) is the electrostatic force, - \(k = 8.9875 \times 10^9 Nm^2C^{-2}\) (Coulomb's constant), - \(q_1 = 1.27 \times 10^{-6} C\) (charge of the fixed bead), - \(q_2 = 6.79 \times 10^{-6} C\) (charge of the sliding bead), and - \(r\) is the distance between the charges.
02

Calculate the gravitational force on the second bead

The gravitational force on the second bead can be calculated using the equation: \( F_{g} = m_2 g \), where \(g=9.81 m/s^2\) is the gravitational acceleration and \(m_2=3.77\times10^{-3}kg\) is the mass of the second bead.
03

Set up an equilibrium equation

Since the electrostatic force between the beads should balance the gravitational force on the second bead when it slides on the wire, we can write the equilibrium equation as: \(\sin(\theta) F_{e} = F_{g}\). Plugging in Coulomb's law and the gravitational force equation, we get: \(\sin(\theta) \left( k \frac{q_1 q_2}{r^2} \right) = m_2 g \)
04

Solve for distance 'd'

Now we can solve for 'r' (since 'r' = 'd' in this case): \(\sin(\theta) \left( k \frac{q_1 q_2}{r^2} \right) = m_2 g \) Solving for 'r', we get: \( r = \sqrt{ k \frac{q_1 q_2}{m_2 g \sin(\theta)} } \) Plugging in the given values: \( r = \sqrt{ 8.9875 \times 10^9 \frac{1.27 \times 10^{-6} (6.79 \times 10^{-6})}{(3.77\times10^{-3})(9.81)\sin(51.3)} }\) Which after solving evaluates to: \(r = 0.0206 m\) Thus, the distance 'd' at which the electrostatic force balances the gravitational force is \(d=0.0206 m\) or \(2.06 cm\).

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