What is the force on an electric dipole due to a line of charge?

[KillerZ]

Homework Statement

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

Homework Equations

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

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

F = F_{on -q} + F_{on +q} = 0

F_{on -q} = -qE
F_{on +q} = +qE

The Attempt at a Solution

I am not sure if I am doing this correct?

F = F_{on -q} + F_{on +q}
F = -qE + +qE

= \frac{pE}{s} + \frac{pE}{s}

= \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}

 

[Redbelly98]

Homework Statement

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

Homework Equations

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

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

F = F_{on -q} + F_{on +q} = 0

F_{on -q} = -qE
F_{on +q} = +qE

The Attempt at a Solution

I am not sure if I am doing this correct?

F = F_{on -q} + F_{on +q}
F = -qE + +qE

= \frac{pE}{s} + \frac{pE}{s}

= \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}

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.