Question

Two 11.5-mg balls are connected by a 28.9-cm-long insulating rod of negligible mass. One ball has a charge of +q and the other has a charge of −q. The system is placed in a 592 N/C uniform electric field. Initially the system is at rest and makes an angle α= 58.9° with respect to the field. It is then released, and when it is momentarily aligned with the electric field, its kinetic energy is 4.86 mJ. What is the magnitude of q?

Answer 1

58.8 μC

Step-by-step explanation:

The two balls create a dipole of dipole moment , p = qd where q = charge and d = distance between the charges = 28.9 cm = 0.289 m. The potential energy change ΔU = -W where W is the work done by the electric field of magnitude E = 592 N/C.

Now U = -pdEcosФ = -qdEcosФ where Ф is the angle between p and E. Since Ф is initially 58.9° and then becomes 0° when p and E align, the potential energy change is thus.

ΔU = -qdEcos0 - (-qdEcos58.9°) = qdE(cos58.9° - 1) = qdE(0.5165 - 1) = -0.4835qdE.

W = - ΔU = -(-0.4835qdE) = 0.4835qdE

This work equals the kinetic energy of the systems. Since K.E = 4.86 mJ = 4.86 × 10⁻³ J.

So K.E = W

K.E = 0.4835qdE

making q subject of the formula

q = K.E/0.4835dE

q = 4.86 × 10⁻³ J/(0.4835 × 0.289 m × 592 N/C)

q = 4.86 × 10⁻³ J/82.72

q = 0.0588 × 10⁻³ C

q = 58.8 × 10⁻⁶ C

q = 58.8 μC