Question

A potassium-chlorine (KCl) molecule has a dipole moment of 8.9 x 10–30 C m. This dipole moment can be thought of as due to a separation of K+ and Cl– by distance d.

a) Calculate d. The charge of K+ is +e, and the charge of Cl– is –e.

b) What is the maximum magnitude of the torque that a uniform electric field with magnitude 3.0 x 104 N/C can exert on a KCl molecule? Show in a sketch the relative orientations of the electric dipole moment p and the electric field E when the torque is a maximum.

c) How much energy is required to reorient the molecule 180o – from p being parallel to E to p being antiparallel to E?

Answer 1

Given:

Dipole moment, P = 8.9*10^-30 Cm.

Charge on K⁺ = 1 e = 1.6*10^-19 C

Charge on Cl^- = 1 e = 1.6*10^-19 C

a) We know that:

Dipole moment, P = q*d

Where q = values of charges

d = distance between the charges

⇒ 8.9*10^-30 = (1.6*10^-19)(d)

⇒ d = 5.56*10^-11 m

b) Torque on a dipole moment, τ = PEsinθ

Where E = Strength of electric field

θ = Angle between the Electric fied vector and dipole

For max Torque, sinθ = 1 (θ = 90°)

⇒ τ = (8.9*10^-30)(3*10⁴)(1)

= 2.67*10^-24 Nm

Diagram:

c) Eergy required to rotate the dipole = Change in its potential energy

= U_f - U_i

= (-PEcosθ_f) - (-PEcosθ_i) Where θ = Angle with the electric field

= -PEcos180° + PEcos0°

= PE + PE

= 2PE = 2*(8.9*10^-30)(3*10⁴)

= 5.34*10^-25 J