anonymous
  • anonymous
Show that an infinite line of charge with linear charge density lamda exerts an attractive force on an electric dipole with magnitude F = (2)(Lamda)(p) / (4)(pie)(Epsilon knot)(r^2). Assume that r is much larger than the charge separation in the dipole.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
http://web.mit.edu/6.013_book/www/chapter11/11.8.html
anonymous
  • anonymous
What part of this is the answer?
anonymous
  • anonymous
I don't understand what the answer is

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anonymous
  • anonymous
start with the field of an infinite line of charge, what is that?
anonymous
  • anonymous
E= 1/(4pi€.) * ( 2(lambda))/r. Then what do I do?
anonymous
  • anonymous
differentiate and multiply by p :)
anonymous
  • anonymous
How would I differentiate? By dx?
anonymous
  • anonymous
did you look over the "force on a dipole" section?
anonymous
  • anonymous
Is the derivation clear?
anonymous
  • anonymous
r
anonymous
  • anonymous
Yes. I think. Lol
anonymous
  • anonymous
Do I differentiate or integrate?
anonymous
  • anonymous
differentiate that upside down triangle is the gradient (space derivative)
anonymous
  • anonymous
here everything only depends on r, no x's y's or z's needed to characterize the problem...
anonymous
  • anonymous
so the gradient is just the derivative with respect to r
anonymous
  • anonymous
Differintiating will get rid of r
anonymous
  • anonymous
nope. r is the variable.
anonymous
  • anonymous
what's the derivative of 1/r with respect to r?
anonymous
  • anonymous
\[-1/r ^{2}\]
anonymous
  • anonymous
Okay I got it. Is the final answer suppose to be negative?
anonymous
  • anonymous
all the rest of the terms are constants, they stay unchanged... multiply by the dipole moment (p) and you're done...
anonymous
  • anonymous
yes negative r hat is towards the center so it's an attractive force...
anonymous
  • anonymous
Oh now it makes sense thank you so much!! I may pass my quiz tomorrow now!
anonymous
  • anonymous
Hope it helped:) gl on the quiz!
anonymous
  • anonymous
Thanks

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