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Force on electric dipole

  • Thread starter KillerZ
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  • #1
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Homework Statement



Show that a line of charge density [tex]\lambda[/tex] exerts an attractive force on an electric dipole with magnitude F = [tex]\frac{2\lambda p}{4\pi\epsilon_{0}r^{2}}[/tex]. Assume that r is much larger than the charge separation in the dipole.

23ti1j5.png



Homework Equations



I need to show that the magnitude, F = [tex]\frac{2\lambda p}{4\pi\epsilon_{0}r^{2}}[/tex]

p = qs => assume s is very small at end of calculations

F = Fon -q + Fon +q = 0

Fon -q = -qE
Fon +q = +qE

The Attempt at a Solution



I am not sure if I am doing this correct?

F = Fon -q + Fon +q
F = -qE + +qE

= [tex]\frac{pE}{s} + \frac{pE}{s}[/tex]

= [tex]\frac{p}{s}\frac{1}{4\pi\epsilon_{0}}\frac{2\lambda}{r} + \frac{p}{s}\frac{1}{4\pi\epsilon_{0}}\frac{2\lambda}{r}[/tex]
 

Answers and Replies

  • #2
Redbelly98
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Homework Statement



Show that a line of charge density [tex]\lambda[/tex] exerts an attractive force on an electric dipole with magnitude F = [tex]\frac{2\lambda p}{4\pi\epsilon_{0}r^{2}}[/tex]. Assume that r is much larger than the charge separation in the dipole.

23ti1j5.png



Homework Equations



I need to show that the magnitude, F = [tex]\frac{2\lambda p}{4\pi\epsilon_{0}r^{2}}[/tex]

p = qs => assume s is very small at end of calculations

F = Fon -q + Fon +q = 0

Fon -q = -qE
Fon +q = +qE

The Attempt at a Solution



I am not sure if I am doing this correct?

F = Fon -q + Fon +q
F = -qE + +qE

= [tex]\frac{pE}{s} + \frac{pE}{s}[/tex]

= [tex]\frac{p}{s}\frac{1}{4\pi\epsilon_{0}}\frac{2\lambda}{r} + \frac{p}{s}\frac{1}{4\pi\epsilon_{0}}\frac{2\lambda}{r}[/tex]
There are 2 problems here:

  • A "-" sign got lost from the -qE term, when you went on to the next step.
  • The distance to the +q charge is not r.

Hope that helps. :smile:
 

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