# Homework Help: Work to rotate a dipole in an electric field

1. Jul 29, 2010

### lachy89

NOTE: BEFORE YOU RESPOND PLEASE KNOW THIS IS AN ASSIGNMENT QUESTION (worth 1/6 % of total subject) AND AS SUCH I DO NOT WANT AN ANSWER TO BE GIVEN TO THIS PROBLEM.

I have tried this question many times in different ways and unfortunately I am unable to get any of these multiple choice answers.

My question to you is: Is one of these answers correct, or has a mistake in the question/ multiple choice answers been made and as such the actual answer is not one of the multiple choices given?

PLEASE DO NOT SHOW WORKING OUT OR GIVE AN ANSWER TO THE QUESTION

Equations that I have used include:

U= - p.E

U= -pE x cos (theta)

W = Ep Intergral of (cos(phi) d(phi)) from pi/3 to 0

1. The problem statement, all variables and given/known data
A dipole with moment 17.0 mC∙m is oriented at an angle of pi/6 above the x-axis. How much work is required to rotate the dipole until it is oriented along the x-axis if the dipole is located in an electric field with strength 87.0 (in positive x-direction) kN/C?

1. -97.5 J
2. 618 J
3. 97.5 J
4. -618 J

Last edited: Jul 29, 2010
2. Jul 29, 2010

### merryjman

I'm responding because no one has yet, even though I'm not exactly sure I did it right. I did a naive calculation and got an answer close to one of the choices, but again I'm not sure if I did it correctly.

3. Jul 29, 2010

### lachy89

I'll show some of my working out in algebraic form so It will be easier to give some feedback.

Torque = -p x E x sin(theta)

Work = Torque x angle (in radians)

Torque in this case is not constant as the torque changes with angle.

Work = SUM OF ( Torque (theta) x d(theta)) from Pi/6 to 0

Work = Integral from Pi/6 to 0 of (Torque (theta) d(theta)

Work = Integral from Pi/6 to 0 of (- E x P x sin (theta) d(theta) )

Work = - E x P x [ -cos (Pi / 6) + cos (0) ]

That does not give me any of the multiple choice answers. Have I made a mistake in my derivation somewhere?