Chapter 24, Problem 24.72CP

Two square conducting plates with sides of length L are separated by a distance D. A dielectric slab with constant K with dimensions L × L × D is inserted a distance x into the space between the plates, as shown in Fig. P24.72. (a) Find the capacitance C of this system, (b) Suppose that the capacitor is connected to a battery that maintains a constant potential difference V between the plates. If the dielectric slab is inserted an additional distance dx into the space between the plates, show that the change in stored energy is

$\text{dU}\text{ = + }\frac{(\text{K}-\text{ 1)}{\in }_{\text{0}}{\text{V}}^{\text{2}}\text{L}}{2\text{D}}\text{dx}$

(c) Suppose that before the slab is moved by dx, the plates are disconnected from the battery, so that the charges on the plates remain constant. Determine the magnitude of the charge on each plate, then show that when the slab is moved dx farther into the space between the plates, the stored energy changes by an amount that is the negative of the expression for dU given in part (b). (d) If F is the force exerted on the slab by the charges on the plates, then dU should equal the work done against this force to move the slab a distance dx. Thus dU = −Fdx. Show that applying this expression to the result of part (b) suggests that the electric force on the slab pushes it out of the capacitor, while the result of part (c) suggests that the force pulls the slab into the capacitor, (e) Figure 24.16 shows that the force in fact pulls the slab into the capacitor. Explain why the result of part (b) gives an incorrect answer for the direction of this force, and calculate the magnitude of the force. (This method does not require knowledge of the nature of the fringing field.)

Figure P24.72