Having Trouble With Dipole Torque and Work

[ccelt09]

There is a two part problem that I'm struggling with.

1) A dipole with a moment of 2.5 nC . m is oriented 33 degrees from a field with strength 3.6 MN/C.

What is the magnitude of the torque on the dipole? ____ mN . m

I know the equation T = qd x E sin (theta) where qd is the given dipole moment, but even with converting and plugging in I don't get the right answer.

2)How much work is required to rotate the dipole until it's antiparallel to the field?

again i tried plugging into -pEcos(theta) but my answer is wrong.I am not quite sure why my approach isn't working or what I am doing wrong. I would greatly appreciate any help.

 

[Doc Al]

Show what you did so we can see what went wrong.

 

[ccelt09]

I used the equation torque = dipole magnitude x E. the dipole came in units 2.5 nC . m, the field as 3.6 MN/C. I took the n to represent nano ^ -9 and M as Mega ^ 6. I did the appropriate conversions to have the product of the two come in units mN.m in which the program wants the answer.

2.5x10^6 mC.m x 3.6x10^6 N/C = 9x10^12 mN.m

2.5 nC x 10^6 to get to 2.5 mC . m and 3.6MN/C x 10^6 to get N/C. I feel that was probably a misinterpretation of what the units represent.

For the second part in order to find the work necessary to make the dipole antiparallel to the field i used W=-pEcos(theta)

-2.5x10^6 mC.m x 3.6x10^6 cos(57) = -4.9x10^12 ; already angled at 33 degrees, 57+33=90 = antiparallel/perpendicular

 

[Doc Al]

I used the equation torque = dipole magnitude x E. the dipole came in units 2.5 nC . m, the field as 3.6 MN/C. I took the n to represent nano ^ -9 and M as Mega ^ 6. I did the appropriate conversions to have the product of the two come in units mN.m in which the program wants the answer.

2.5x10^6 mC.m x 3.6x10^6 N/C = 9x10^12 mN.m

2.5 nC x 10^6 to get to 2.5 mC . m and 3.6MN/C x 10^6 to get N/C. I feel that was probably a misinterpretation of what the units represent.

You're messing up the units a bit:
2.5 nC . m = 2.5x10^-6 mC.m (not 2.5x10⁺⁶).

For the second part in order to find the work necessary to make the dipole antiparallel to the field i used W=-pEcos(theta)

-2.5x10^6 mC.m x 3.6x10^6 cos(57) = -4.9x10^12 ; already angled at 33 degrees, 57+33=90 = antiparallel/perpendicular

That equation tells you the potential energy, U = -pEcos(theta). To find the work, figure out the change in PE.

 

[rwisz]

I understand your struggle with dipole torque and work. It can be a complex topic to grasp and apply correctly. Let's break down the two parts of the problem and discuss some potential solutions.

1) To find the magnitude of the torque on the dipole, we can use the equation T = qd x E sin(theta), as you mentioned. However, it is important to note that the dipole moment, qd, should be in units of Coulomb-meters (C-m), not nanocoulomb-meters (nC-m). This could be the reason why your answer is incorrect. So, let's convert the dipole moment to C-m by multiplying it by 10^-9. This gives us a dipole moment of 2.5 x 10^-9 C-m.

Now, we can plug in our values into the equation: T = (2.5 x 10^-9 C-m) x (3.6 x 10^6 N/C) x sin(33 degrees). This gives us a torque of approximately 2.55 x 10^-2 mN-m.

2) For the second part, we need to find the work required to rotate the dipole until it is antiparallel to the field. The equation for work in this case is W = -pEcos(theta), where p is the magnitude of the dipole moment. Again, we need to make sure that the dipole moment is in units of C-m. So, we can use the same conversion as before to get a dipole moment of 2.5 x 10^-9 C-m.

Next, we need to find the angle at which the dipole is antiparallel to the field. Since the initial angle is 33 degrees, the final angle would be 180 degrees - 33 degrees = 147 degrees. So, we can plug in our values into the equation: W = -(2.5 x 10^-9 C-m) x (3.6 x 10^6 N/C) x cos(147 degrees). This gives us a work of approximately -1.60 x 10^-8 J.

It is possible that your answer was incorrect because you did not use the correct units for the dipole moment, or you may have used the wrong angle in the cosine function. I hope this explanation helps you in solving similar problems in the future. Keep