## Physics Friday 11

Suppose we have a parallel plate capacitor with a distance d between square plates of width w and length L. Along the length L, we partially insert a dielectric of thickness d, width w, and dielectric constant k>1; we indicate the length of inserted dielectric as x. If there is a charge on the plates, the dielectric will experience a force inward. Neglecting edge effects and the dielectric constant of air, what is the force on the dielectric for:
(1) a constant charge Q on the capacitor plates (one has +Q, the other -Q)?
(2) a constant potential difference V (i.e. the capacitor is connected to an ideal voltage source)?

To solve, we treat the portion of the capacitor with the dielectric and the portion without as capacitors in parallel; one filled with a dielectric of constant k and plate area xw, the other filled with air (kair=1 for this treatment) and with plate area (L-x)w. Using the formula for a parallel plate capacitor neglecting edge effects , we find:




To compute the force F, we first compute the energy U stored in the capacitor. If the force F acting inward moves the dielectric inward by dx, then the force has done work F•dx, and so the energy must have decreased by an amount dU equal in magnitude to this: dU=-F•dx, and thus .
Now, the energy stored in a capacitor is given by .

(1) Force for constant charge:
Here, we have the charge is constant, so the formula we use for the energy stored is .

Thus 

The force varies with x, decreasing as x increases.

(2) Force for constant electric potential:
Here the formula we use for the energy stored is 

Thus 

Note here that the force is independent of x.

Note: in both cases, we see that the formulas we obtained for the force does not become zero at x=L (fully inserted dielectric). This is because we neglected edge effects. However, as long as L-x is not small (and the usual conditions for small edge effects hold) the above serve as reasonable approximations.