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.

 

[thomate1]

= \frac{2\lambda p}{4\pi\epsilon_{0}r^{2}}

First, we can rewrite the electric dipole moment, p, as p = qs, where q is the magnitude of the charges and s is the separation between them. Since we are assuming that the distance r is much larger than the separation s, we can neglect the term s in our calculation.

Next, we can use the electric field due to a line of charge, E = \frac{\lambda}{2\pi\epsilon_{0}r}, to calculate the forces on the individual charges in the dipole. The force on the negative charge, Fon -q, is equal to -qE, while the force on the positive charge, Fon +q, is equal to +qE.

Adding these two forces together, we get:

F = Fon -q + Fon +q = -qE + +qE = \frac{pE}{s} + \frac{pE}{s}

Since we are neglecting the separation s, this becomes:

F = \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}

Finally, we can simplify this expression to get:

F = \frac{2\lambda p}{4\pi\epsilon_{0}r^{2}}

This is the same expression given in the homework statement, showing that the force on an electric dipole due to a line of charge is attractive and has a magnitude of \frac{2\lambda p}{4\pi\epsilon_{0}r^{2}}.