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?

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