Question

Object A has mass 83.0 g and hangs from an insulated thread. When object B, which has a charge of +140 nC, is held nearby, A is attracted to it. In equilibrium, A hangs at an angle θ = 7.20° with respect to the vertical and is 5.40 cm to the left of B.
a. What is the charge on A?
b. What is the tension in the thread?

Answer 1

Answer:

a) -238 nC  

b) 0.889 N  

Explanation:

Concepts and Principles

Particle in Equilibrium: If a particle maintains a constant velocity (so that a = 0), which could include a velocity of zero, the forces on the particle balance and Newton's second law reduces to:  

∑F = 0                                                                           (1)  

Coulomb's Law: the magnitude of the electrostatic force exerted by a point charge q1 on a second point charge q2 separated by a distance r is directly proportional to the product of the two charges and is inversely proportional to the square of the distance between them:

F_12 = k*| q1 |*| q2 |/r^2                                                 (2)

where k = 8.99 x 10^9 N  m^2/C^2 is Coulomb constant.  

Given Data  

mA (mass object A) = (83 g)*(1/1000g)=0.09 kg

qB (charge of object B) = (140 nC)*(1/10^9 nC) = 130 x 10^-9 C

Object A is attracted to object B.

Ф(angle made by object A with the vertical) = 7.2°  

(  r (distance between the two objects) = (5 cm) * (1 m/ 100 cm) =0.05 m  

Object A is in equilibrium.  

Required Data

In part (a), we are asked to determine the charge qA of object A.

In part (b), we are asked to determine the tension T in the thread.  

(a) The FBD in Figure 1 shows the forms acting on object A; Fe is the electric force exerted on object A by object B, T is the tension force exerted on the thread, and m_a*g is the gravitational force exerted on object A.  

Model object A as a particle in equilibrium in the horizontal and vertical direction and apply Equation (1) to it:  

∑F_x = F_e-Tsin = 0                                   F_e=TsinФ                (3)

∑F_y = TcosФ - m_a*g= 0                      m_a*g=TsinФ                (4)

Divide Equation (3) by Equation (4) to eliminate T:

F_e/m_a*g=tanФ

F_e=m_a*g*tanФ

Substitute for  F_e by using Coulomb's law from Equation (2):

k*| q_A |*| q_B |/r = m_a*g*tanФ

Solve for q_A :  

| q_A | = m_a*g*tanФ_r/k*| q_B |

Substitute numerical values from given data:

| q_A | =  238 nC  

Because object A is attracted to object B. it has an opposite negative charge. Therefore, the charge on object A is | q_A | =  -238 nC  

(b)  

Solve Equation (4) for T:  

T = m_a*g/cosФ

Substitute numerical values from given data:

T = (0.09 kg)(9.8 m/s^2) /cos 7.2°  

  = 0.889 N