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Nonuniform electric field and a dipole 
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#1
Feb2109, 01:32 PM

P: 115

1. The problem statement, all variables and given/known data
a dipole, [tex] \vec p [/tex], is in a non uniform field, [tex] \vec E = (E) \vec i [/tex] where [tex] \vec E [/tex] points along the x axis. What is the net force on the dipole if [tex] \vec E [/tex] depends only on x? 2. Relevant equations [tex] \vec{p} \times \vec{E} =  \vec p   \vec E  sin \theta [/tex] 3. The attempt at a solution i think if the charges are separated by dx, [tex] \vec p = (Q) (dx) (\vec i) [/tex] then, i get a bit confused.... what thinking step shall I take next??? 


#2
Feb2109, 08:54 PM

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#3
Feb2109, 09:02 PM

P: 115

[tex] Q \vec E = \vec F [/tex] pure dipole. 


#4
Feb2109, 09:05 PM

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Nonuniform electric field and a dipole



#5
Feb2109, 09:14 PM

P: 115

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! [tex] \vec F = \vec F_+ + \vec F_ = (Q \vec E_+  Q \vec E_ ) \cdot \vec i [/tex] what thinking step shall I take next??? 


#6
Feb2109, 09:27 PM

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The general equation for the force on a dipole (pure or physical) is derived in almost every introductory level EM 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
Feb2109, 09:35 PM

P: 115




#8
Feb2109, 09:43 PM

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If so, then you can continue on from here: 


#9
Feb2209, 02:50 AM

P: 115

if the dipole vector [tex] \vec p [/tex] is parallel to the x axis and points to the + x direction, then
[tex] \vec E = (E) \vec x [/tex] [tex] \vec F_ =  ( \vec E ) (Q) [/tex] [tex] \vec F_+ = ( \vec E ) (Q) [/tex] [tex] \vec F_ =  (E) \vec x (Q) [/tex] [tex] \vec F_+ = (E) \vec x (Q) = (E) (x + dx) (Q) [/tex] i'm stuck... 


#10
Feb2209, 03:33 AM

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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 [itex]\pm Q[/itex] separated by a distance [itex]dx[/itex], (2) the second is that the dipole points in the xdirection, (3) and the third is that the negative charge is located at [itex]\vec{r}=x\hat{i}[/itex] From your original problem statement, I see absolutely no reason to make these assumptions: 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
Feb2209, 04:01 AM

P: 115

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
Feb2209, 04:12 AM

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P: 5,003

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 [tex]\vec{F}(\vec{r})=(\vec{p}\cdot\hat{i})E'(x)\hat{i}[/tex] where [itex]\vec{r}[/itex] is the location of the dipole) 


#13
Feb2209, 07:39 AM

Sci Advisor
P: 1,261

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


#14
Feb2209, 11:59 AM

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[tex]\vec{F}=(\vec{p}\cdot\vec{\nabla})\vec{E}[/tex] 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
Feb2209, 07:40 PM

P: 115

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
Feb2209, 07:54 PM

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P: 5,003

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
Feb2209, 07:57 PM

P: 115

Thanks gab,
are you a grad student or a practicing engineer? 


#18
Mar1909, 08:31 AM

P: 1

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 Alevel 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: [tex]{\vec F}=\nabla({\vec p\cdot{\vec E}}).[/tex] This is incorrect, gabbagabbahey posted the correct version. If you compare the component [tex]F_x[/tex] for the two answers you'll see they are inconsistent. Vector calculus... 


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