# Force on electric dipole

## 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 = 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

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

= $$\frac{p}{s}\frac{1}{4\pi\epsilon_{0}}\frac{2\lambda}{r} + \frac{p}{s}\frac{1}{4\pi\epsilon_{0}}\frac{2\lambda}{r}$$

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Redbelly98
Staff Emeritus
Homework Helper

## 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 = 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

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

= $$\frac{p}{s}\frac{1}{4\pi\epsilon_{0}}\frac{2\lambda}{r} + \frac{p}{s}\frac{1}{4\pi\epsilon_{0}}\frac{2\lambda}{r}$$
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. 