- #1
BigTanker
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Homework Statement
A point charge, Q, is "nailed down" on a table. Around it, at radius R, is a frictionless circular track on which a dipole [itex]\boldsymbol{p}[/itex] rides, constrained to always point tangent to the circle. Show that the electric force on the dipole is (in the forward direction of the dipole):[tex]\boldsymbol{F} = \frac{Q}{4 \pi \epsilon_0} \frac{\boldsymbol{p}}{R^3}[/tex]
Homework Equations
[tex]\boldsymbol{F} = ( \boldsymbol{p} \cdot \nabla ) \boldsymbol{E}[/tex]
The Attempt at a Solution
I started by recognizing that the electric field from the point charge is [tex]\boldsymbol{E} = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2} \hat{r}.[/tex]
Thus, I feel like the force should be: [tex]\boldsymbol{F} = ( \boldsymbol{p} \cdot \nabla ) \boldsymbol{E} = \boldsymbol{p} \frac{-2}{4 \pi \epsilon_0} \frac{Q}{R^3} = \frac{-Q}{2 \pi \epsilon_0} \frac{\boldsymbol{p}}{R^3}.[/tex]
However, if I draw the dipole out as 2 separated point charges, +q and -q, I see that the net force should point in the direction as the dipole, not opposite the direction of the dipole (as my math above would suggest).
Where have I erred?