Question

The separation between the oxygen and carbon atoms in a carbon molecule is 0.113 nm. Assume the charges are +1.0e and -1.0e.

a) Find the electric dipole moment of the CO molecule.

b) We immerse the molecule into a uniform electric field of strength 2.0 x 10⁵ N/C and oriented such that the electric dipole moment is making an angle of 53 degrees with respect to the field. What is the net torque in the molecule around its center of mass?

c) If released from rest, what is the initial angular acceleration of the molecule?

d) What are the inital linear accelerations of the carbon and oxygen atoms, respectively?

Answer 1

Data Given

q₁ =q₂ = 1.0 e

distance d = 0.113 nm, E = 2.0*10⁵ N/C, angle = 53⁰

Part A) Electric dipole moment of a dipole is given by the product of either charge to distance beetween the charges or length of the dipole.

$p = q.d = 1.0e\times 0.113 nm = 1.6\times 10^{-19}C\times 0.113\times 10^{-9}m$

$p = 1.808\times 10^{-29}Cm$

Part B) Torque experienced by a dipole in a uniform electric field is given by

$\vec{\tau } = \vec{p}\times \vec{E}$

$\tau = pE\sin \theta$

$\tau = 1.808\times 10^{-29}Cm\times 2\times 10^{5}N/C\sin 53^{0}$

$\tau = 2.89\times 10^{-27}Nm$

Part C) Moment of inertia of a CO molecule about its centre is given by

$I = \mu d^{2}$   where

$\mu = \frac{m_{C}m_{O}}{(m_{C}+m_{O})N_{A}}=\frac{0.012\times 0.016}{(0.012+0.016)6.023\times 10^{23}}= 1.139\times 10^{-26}kg$

$I = 1.139\times 10^{-26}\times (0.113\times 10^{-9})^{2} kg.m^{2}$

$I = 14.5\times 10^{-47} kg.m^{2}$

We know that

$\tau= I\alpha = 2.89\times 10^{-27}Nm$

$\alpha = \frac{2.89\times 10^{-27}Nm}{14.5\times 10^{-47} kg.m^{2}} = 1.993\times 10^{19}rad/s^{2}$

Part D) Linear acceleration of the system, we know that

$a = \alpha d = 1.993\times 10^{19}rad/s^{2}\times 0.113\times 10^{-9} m = 2.25\times 10^{9}m/s^{2}$