# Non-uniform electric field and a dipole

1. Feb 21, 2009

### orthovector

1. The problem statement, all variables and given/known data
a dipole, $$\vec p$$, is in a non uniform field, $$\vec E = (E) \vec i$$ where $$\vec E$$ points along the x axis. What is the net force on the dipole if $$\vec E$$ depends only on x?

2. Relevant equations

$$| \vec{p} \times \vec{E}| = | \vec p | | \vec E | sin \theta$$

3. The attempt at a solution

i think if the charges are separated by dx, $$\vec p = (Q) (dx) (\vec i)$$

then, i get a bit confused.... what thinking step shall I take next???

Last edited: Feb 21, 2009
2. Feb 21, 2009

### gabbagabbahey

This equation gives the torque on a physical dipole (about its center) in a uniform field, or the torque on a pure (ideal) dipole about its center, not the force on a dipole in a non-uniform field. Look in your textbook to find the equation for force on a dipole (physical or ideal) in a non-uniform field.

You seem to be treating the dipole as a physical dipole....are you told whether it is a physical dipole or pure dipole?

3. Feb 21, 2009

### orthovector

yes, I didn't write down the equation for the force.

$$Q \vec E = \vec F$$
pure dipole.

4. Feb 21, 2009

### gabbagabbahey

No, that's the force on a monopole (point charge) in any electric field......read your textbook!

5. Feb 21, 2009

### orthovector

oh come on, you're treating me like some scrub. I know the equation... positive force plus negative force in which the Electric field from the dipole decreases by 1/r cubed for far distances....

I've done a lot of thinking on this problem..

GeeZ!

$$\vec F = \vec F_+ + \vec F_- = (Q \vec E_+ - Q \vec E_- ) \cdot \vec i$$

what thinking step shall I take next???

6. Feb 21, 2009

### gabbagabbahey

For a physical dipole, with two charges $\pm Q$ separated by some non-zero distance, then this would be a good start, but you just said that you have a pure dipole, not a physical dipole.

The general equation for the force on a dipole (pure or physical) is derived in almost every introductory level E-M textbook I've come across...is it truly not derived in your textbook?....Which textbook does your course use and what level course is this for?

7. Feb 21, 2009

### orthovector

sorry, it must be a physical dipole. The two charges are separated by a distance dx.

8. Feb 21, 2009

### gabbagabbahey

Okay, and you are told that the charges are $\pm Q$?

If so, then you can continue on from here:

Where are the charges located? What does that make $E_{\pm}$?

9. Feb 22, 2009

### orthovector

if the dipole vector $$\vec p$$ is parallel to the x axis and points to the + x direction, then

$$\vec E = (E) \vec x$$
$$\vec F_- = - ( \vec E ) (Q)$$
$$\vec F_+ = ( \vec E ) (Q)$$
$$\vec F_- = - (E) \vec x (Q)$$
$$\vec F_+ = (E) \vec x (Q) = (E) (x + dx) (Q)$$

i'm stuck...

Last edited: Feb 22, 2009
10. Feb 22, 2009

### gabbagabbahey

What a jumbled mess!....please use \hat{} to represent unit vectors and \vec{} for normal vectors; it will make things less confusing....and be careful with your brackets! ($(E) \vec x$ means E times the vector x; while $E(x)\hat{i}$ means E as a function of x, directed along the x-axis--which is what I'm sure you meant to write)

I'm sorry if this sounds rude, but I think you should consider getting a tutor. I think some face to face assistance could really benefit you.

At this point, you seem to have made at least 3 assumptions without justification; (1) the first is that the dipole consists of two point charges $\pm Q$ separated by a distance $dx$, (2) the second is that the dipole points in the x-direction, (3) and the third is that the negative charge is located at $\vec{r}=x\hat{i}$

From your original problem statement, I see absolutely no reason to make these assumptions:
Is this how the original problem is worded on your assignment sheet/textbook?

Again I ask what level course this is for, and what the course textbook is? (If I have a copy of the same text, I can point to any helpful sections)

11. Feb 22, 2009

### orthovector

That was the entire problem. the 3 assumptions I made are all valid. (1) 2 point charges separated by a distance d is a useful summary to simplify any dipole system. (2) the dipole vector can be anti parallel, parallel, or orthogonal to the Electric field, but since the electric field is a function of x, the dipole will eventually orient itself in a parallel position to the Electric field, and it will eventually move horizontally due to the difference in forces of the negative charge and the positive charge. (3) this part is already explained.

I'm offended by your remark. It seems like you haven't put any thought into this problem, or maybe you don't understand...and you say I made unjustified claims???

12. Feb 22, 2009

### gabbagabbahey

Yes and no. The limiting process necessary to get results that are consistent with a pure dipole (which is what is typically meant when you are told the you have a dipole $$\vec{p}$$ and not specifically told whether it is pure or physical) is fairly complicated. You have to take the limit as d-->0 and Q-->infinity while holding the product Qd equal to p. You can do the problem this way if you so choose, but there is a much simpler method.

In electrostatics one typically analyzes a charge distribution at a particular snapshot in time. I see no reason to assume that the questioner wants you to analyze the moment in time where p is aligned in the x-direction and has moved some distance away from its origin.

Instead, I would assume that p points in any direction (that way you can derive what the force on the dipole is in the general case, instead of just when it is aligned with the field). I would also assume that the question refers to a pure dipole, and that E=E(x)i.

In such a case, it is easy to show that $$\vec{F}(\vec{r})=(\vec{p}\cdot\hat{i})E'(x)\hat{i}$$ where $\vec{r}$ is the location of the dipole)

Last edited: Feb 22, 2009
13. Feb 22, 2009

### clem

At the risk of entering such a contentious thread, I will inform you that the force on a dipole $${\vec p}$$ in an electric field $${\vec E}$$ is
$${\vec F}=\nabla({\vec p\cdot{\vec E}}).$$

Last edited: Feb 22, 2009
14. Feb 22, 2009

### gabbagabbahey

For a pure dipole, yes. In the more general case,

$$\vec{F}=(\vec{p}\cdot\vec{\nabla})\vec{E}$$

This is the formula that I recommended he look up in his textbook, as it is derived in almost every introductory level EM text I've come across.

15. Feb 22, 2009

### orthovector

thanks gab,

My introductory physics book does not have this derivation. Instead, it asks me to figure out the derivation. Would you recommend a better physics textbook that has this derivation? REZNICK? GIANCOLI? IDA? SERWAY?? KNIGHT? WHICH ONE???

16. Feb 22, 2009

### gabbagabbahey

I don't have those texts handy right now, but I seem to recall seeing the derivation in Serway, and it can certainly be found in intermediate level texts Griffiths' Introduction to Electrodynamics and Jackson's Electrodynamics

However, seeing as it isn't in your text, you must be expected to derive the expression in post #13 by means of a Taylor series expansion and the limiting process I described at the beginning of the post.

Try expanding the E(x+dx) around the point x assuming dx<<x and then substitute that into your expression for the force. (since you don't know what E(x) is, there is no way to dtermine what E'(x), E''(x) etc. are so just write them as E'(x) etc.)

17. Feb 22, 2009

### orthovector

Thanks gab,

are you a grad student or a practicing engineer?

18. Mar 19, 2009

### aphexTriplet

I appreciate this thread is inactive but in case someone stumbles upon it...

Books:
Electricity and Magnetism - Duffin
Electricity and Magnetism - Bleaney & Bleaney

Are both good books suitable for first year undergrads, in fact Duffin is probably accessible to most A-level students. Someone above recommended Jackson's Electrodynamics which is more of an advanced reference text really, the books listed above are a better place to start.

Clem posted the force on an idealised dipole being:
$${\vec F}=\nabla({\vec p\cdot{\vec E}}).$$

This is incorrect, gabbagabbahey posted the correct version. If you compare the component $$F_x$$ for the two answers you'll see they are inconsistent. Vector calculus...