Question

Two balls have the same mass, \(0.9935 \mathrm{~kg}\), and the same charge. They hang from the ceiling on strings of identical length, \(\ell=1.235 \mathrm{~m}\), as shown in the figure. The angle of the strings with respect to the vertical is \(22.35^{\circ}\) What is the charge on each ball?

Short answer

Question: Calculate the charge on each identical conducting ball in a static equilibrium system, given the mass of each ball is 0.9935 kg, the length of the strings suspending the balls is 1.235 m, and the angle the strings make with the vertical is 22.35 degrees. Answer: The charge on each ball is approximately \(2.86 \times 10^{-6} \mathrm{~C}\).

Step-by-step solution

Draw a free body diagram and identify the forces acting on the balls

For each ball, there are three forces acting upon it: the tension force in the string \(T\), the gravitational force \(F_g = mg\), and the electrostatic force \(F_e\) between the charged balls. Draw a free body diagram and label these forces for each ball.

Decompose the forces into horizontal and vertical components

By decomposing the forces acting on each ball into their horizontal and vertical components, we can analyze them in terms of the angle made by the strings. - For the tension force, we have the horizontal component \(T_x = T\sin(\theta)\) and the vertical component \(T_y = T\cos(\theta)\). - The gravitational force acts vertically downward, so \(F_{g_y} = mg\). - The electrostatic force acts horizontally, pushing the balls apart, so \(F_{e_x} = F_e\).

Set up the static equilibrium equations for the horizontal and vertical forces

In static equilibrium, the net force acting on each ball in both the horizontal and vertical directions should be zero. Hence, we can set up the following equations: 1. For the horizontal forces: \(T_x = F_e\) 2. For the vertical forces: \(T_y = mg\) Now substitute the expressions from Step 2 into the equations: 1. \(T\sin(\theta) = F_e\) (Equation 1) 2. \(T\cos(\theta) = mg\) (Equation 2)

Solve for the electrostatic force

Divide Equation 1 by Equation 2 to eliminate T: $$\frac{\sin(\theta)}{\cos(\theta)} = \frac{F_e}{mg}$$ Now, solve for \(F_e\): $$F_e = mg\tan(\theta)$$ Plug in the given values for mass (\(m = 0.9935 \mathrm{~kg}\)) and angle (\(\theta = 22.35^{\circ}\)), as well as the acceleration due to gravity (\(g = 9.81 \mathrm{~m/s^2}\)) to find \(F_e\). $$F_e = (0.9935 \mathrm{~kg})(9.81 \mathrm{~m/s^2})\tan(22.35^{\circ})$$ $$F_e \approx 4.167 \mathrm{~N}$$

Solve for the charge using Coulomb's Law

The electrostatic force between the charged balls can be calculated using Coulomb's Law: $$F_e = \frac{kq^2}{r^2}$$ Where \(k = 8.99 \times 10^9 \mathrm{~N \cdot m^2/C^2}\) is the electrostatic constant, \(q\) is the charge on each ball, and \(r\) is the distance between the balls. Since the strings have the same length and make the same angle, we can use geometry to determine \(r\): $$r = 2\ell\sin(\theta)$$ Plug in the given values for the length, \(\ell = 1.235 \mathrm{~m}\), and the angle, \(\theta = 22.35^{\circ}\), to find \(r\). $$r = 2(1.235 \mathrm{~m})\sin(22.35^{\circ})$$ $$r \approx 0.961 \mathrm{~m}$$ Now, rearrange Coulomb's law equation to solve for \(q\): $$q = \sqrt{\frac{F_e r^2}{k}}$$ Plug in the calculated values for \(F_e\) and \(r\), as well as the constant \(k\): $$q = \sqrt{\frac{(4.167 \mathrm{~N})(0.961 \mathrm{~m})^2}{8.99 \times 10^9 \mathrm{~N \cdot m^2/C^2}}}$$ $$q \approx 2.86 \times 10^{-6} \mathrm{~C}$$ So the charge on each ball is approximately \(2.86 \times 10^{-6} \mathrm{~C}\).