Question

An electric dipole has opposite charges of 6.63?10?15C separated by a distance of 0.63mm. It is oriented at 68.0

Answer 1

dipole moment p = q*d = 6.63?10?15*0.63*10^-3 = 4.1769*10^-18

Torque = E*p*sin(theta) = 10*10^3*4.1769*10^-18*sin68

= 3.8727^10-14 N-m

Answer 2

dipole moment p = q*d = 6.63?10?15*0.63*10^-3 = 4.1769*10^-18

Torque = E*p*cos(theta) = 10*10^3*4.1769*10^-18*cos68 = 1.565^10-14 N-m

Answer 3

TAKE HELP FROM

1. An electric dipole has opposite charges of 5.0 X 10-15
C which are separated by a distance of 0.50 mm. It is
oriented in a uniform electric field of 1.0 X 103
N/C. If the magnitude of the torque exerted on the dipole by the
electric field is 1.05 X 10-15
Nm, what is its orientation with respect to the electric field?

Position a coordinate system with the center of the dipole at the origin, the positive charge on the x-axis at 0.25 mm, and the negative charge on the negative x-axis at -0.25 mm. Now let's figure out what electric field component in the y direction ONLY, would be required, to exert a torque of 1.05 x 10^(-15) N-m. You can see that an electric field oriented in the +y direction would exert a torque, because it would repel the positive charge while attracting the negative charge, so it would tend to cause the dipole to rotate around the origin towards an orientation parallel to the field lines.

The magnitude of the force that such a field would exert on each charge is |E| times 5.0 x 10^(-15) C. If you think of the dipole as a dumbbell, there are total forces of |E| x 10^(-14)C tending to make it rotate, and these forces provide a torque of FXr = |E| x 10(-14)C x (2.5 x 10^(-4) m)
= 2.5 |E| x 10^(-18) C-m.
If the |E| were 1000 N/C,
the torque would be 2.5 x 10^(-15)Nm.
Since the torque is "only" 1.05 x 10^(-5) Nm,
apparently the component of the E field in the "y" direction is only 1.05/2.5 of its total magnitude, so the angle between the E field and the x-axis (in "my" scheme) is arcsin(1.05/2.5) = 24.8 degrees. This is the angle between the E field and the axis of the dipole.

Answer 4

tau = P x E

=>tau = qd x E x sin?

=>tau = qd x E x sin68*

=> tau = 3.87*10^-14 N-m

Answer 5

Torque T on dipole is given by =LQE cos theta

where L = length og the dipole = 0.63mm

Q = charge = 6.63*10^-15

E = EF = 10*10^4 N/C

T = 0.63*10^-3 * 6.63*10^-15 * 10*10^3 sin 68

T = -3.87 *10^-14 Nm

- sign indicates that motion is counter clock wise