Question

Two spheres, each of mass \(M=2.33 \mathrm{~g}\), are attached by pieces of string of length \(L=45.0 \mathrm{~cm}\) to a common point. The strings initially hang straight down, with the spheres just touching one another. An equal amount of charge, \(q\), is placed on each sphere. The resulting forces on the spheres cause each string to hang at an angle of \(\theta=10.0^{\circ}\) from the vertical. Determine \(q,\) the amount of charge on each sphere.

Short answer

Based on the given scenario of the charged spheres hung by a string and the step by step solution provided, provide a short answer to the problem: The amount of charge on each sphere is approximately \(2.27*10^{-7}\) Coulombs.

Step-by-step solution

Draw a diagram and identify the forces

Begin by drawing a diagram of the system. Label the masses, angles, and forces acting on the spheres. There are three forces acting on each sphere: the electrostatic force between the spheres (\(F_c\)), the gravitational force (\(F_g\)), and the tension in the string (\(T\)).

Identify forces using Newton's second law

By Newton's second law (\(F=ma\)), we can write for sphere 1, \(\Sigma F_x = T\sin(\theta) - F_c =0\) \(\Sigma F_y = T\cos(\theta)-F_g=0\)

Coulomb's Law and Gravitational Force

Now we need to find the electrostatic force \(F_c\) and gravitational force \(F_g\). We can use Coulomb's Law to find the electrostatic force: \(F_c = \frac{kq^2}{r^2}\) where \(k = 8.99 * 10^9 Nm^2/C^2\) (Coulomb's constant), \(q\) is the charge, and \(r\) is the distance between the two spheres. The gravitational force can be found by \(F_g = Mg\), where \(M=2.33 * 10^{-3} kg\) is the mass of each sphere and \(g = 9.81 m/s^2\) is the acceleration due to gravity. Then, \(F_g=(2.33*10^{-3}kg)(9.81m/s^2)=2.28293*10^{-2}N\)

Equilibrium conditions

In equilibrium conditions, the tension in the string can be found from \(T\cos(\theta)=F_g\). Plugging in the values, we get \(T\cos(10^{\circ})=2.28293*10^{-2}N\) This allows us to find the horizontal component of the tension force: \( T\sin(10^{\circ})=F_c\) As we know the angles, and our aim is to find \(q\), we will reformulate \(F_c\): \(T\sin(10^{\circ})=\frac{kq^2}{r^2}\)

Solve for the charge q

Now we can substitute \(T\cos(10^{\circ})\) from step 4 into step 5 equation and solve for the charge q: \(\frac{2.28293*10^{-2}N}{\cos(10^{\circ})}\sin(10^{\circ})=\frac{kq^2}{r^2}\) \(r=L\sin(10^{\circ})\) After substituting values for \(k\), \(F_g\) and \(L\) into the equation, we get: \(\frac{2.28293*10^{-2}N}{\cos(10^{\circ})}\sin(10^{\circ})=\frac{8.99*10^{9}Nm^2/C^2*q^2}{[45*10^{-2}\sin(10^{\circ})]^2}\) Now, solve for \(q\): \(q^2 = \frac{(2.28293*10^{-2}N)(45*10^{-2} \sin(10^{\circ}))^2\cos(10^{\circ})}{8.99*10^9 Nm^2/C^2\sin(10^{\circ})}\) \(q^2 = 5.155*10^{-13}C^2\) Taking the square root, we find the charge on each sphere: \(q = \sqrt{5.155*10^{-13}C^2}= 2.27*10^{-7}C\) After calculating, the amount of charge on each sphere is approximately \(2.27*10^{-7}\) Coulombs.