Question

Three \(5.00-\mathrm{g}\) Styrofoam balls of radius \(2.00 \mathrm{~cm}\) are coated with carbon black to make them conducting and then are tied to \(1.00-\mathrm{m}\) -long threads and suspended freely from a common point. Each ball is given the same charge, \(q\). At equilibrium, the balls form an equilateral triangle with sides of length \(25.0 \mathrm{~cm}\) in the horizontal plane. Determine \(q\)

Short answer

Based on the equilibrium condition and the formula for the electrostatic force between charged objects, the charge on each of the three Styrofoam balls forming an equilateral triangle is approximately \(1.49 \times 10^{-8} \mathrm{C}\).

Step-by-step solution

Analyze the forces acting on each ball

Each ball has three forces acting on it: 1. Gravitational force (F_g): downward acting force due to the mass of the ball. 2. Tension force (T): force exerted by the thread on the ball. 3. Electrostatic force (F_e): force exerted by other Styrofoam balls on the ball due to their charges. Since the balls are in equilibrium, the sum of the forces acting on each ball must be zero.

Determine the gravitational force acting on each ball

We can find the gravitational force acting on each ball using the formula: \(F_{g} = mg\) where m is the mass of the ball (5 g) and g is the gravitational acceleration (9.81 m/s²). First, convert the mass to kilograms: \(m = 5.00 \mathrm{~g} \times \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} = 0.005 \mathrm{~kg}\) Now, calculate the gravitational force: \(F_{g} = (0.005 \mathrm{~kg})(9.81 \mathrm{~m/s^2}) = 0.04905 \mathrm{~N}\)

Determine the angle between the tension force and the vertical axis

Let the triangle be denoted by vertices A, B, and C, where A, B, and C are Styrofoam balls. Let the common point to which the balls are tied be O. We can determine the angle θ between the tension force and the vertical axis by considering the triangle formed by points O, A, and the midpoint of line segment BC. Since triangle ABC is equilateral, the distance from O to the midpoint of BC is half this side length: \(\frac{0.25 \mathrm{~m}}{2}=0.125 \mathrm{~m}\). Knowing that the thread's length is 1.00 m, we can now find the angle θ using the inverse tangent (arctan) function: \(\theta = \arctan{\frac{0.125 \mathrm{~m}}{1.00 \mathrm{~m}}} \approx 7.125 ^{\circ}\)

Calculate the electrostatic force between charged balls

The electrostatic force (F_e) between charged objects can be calculated using Coulomb's law: \(F_{e} = \frac{kq^2}{r^2}\) where k is the electrostatic constant (\(8.99 \times 10^{9} \mathrm{~N\cdot m^2/C^2}\)), q is the charge on each ball, and r is the distance between the balls (0.25 m).

Calculate the equilibrium condition for the forces

At equilibrium, the sum of forces along the vertical and horizontal directions should be zero. For the vertical direction, we have: \(T \sin{\theta} = F_{e}\) For the horizontal direction, we have: \(T \cos{\theta} = F_{g}\) Divide the vertical equation by the horizontal equation to eliminate T: \(\frac{F_{e}}{F_{g}} = \tan{\theta}\)

Solve for the charge q

We know the values for \(F_{g}\) and \(\theta\), so we can substitute those into the equation: \(\frac{\frac{kq^2}{r^2}}{F_{g}} = \tan{\theta}\) Now, isolate the charge q and substitute the known values of k, r, F_g, and θ: \(q^2 = \frac{r^2 F_{g} \tan{\theta}}{k}\) \(q^2 = \frac{(0.25 \mathrm{~m})^2(0.04905 \mathrm{~N})(\tan{7.125 ^{\circ}})}{8.99 \times 10^{9} \mathrm{~N\cdot m^2/C^2}}\) Calculating the value of q^2, then taking the square root: \(q \approx 1.49 \times 10^{-8} \mathrm{C}\) So, the charge on each Styrofoam ball is approximately \(1.49 \times 10^{-8} \mathrm{C}\).