# 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.

Related Introductory Physics Homework Help News on Phys.org

#### Doc Al

Mentor
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

Mentor
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.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving