-Q +Q -Q 15 +Q d' 1) A parallel plate capacitor (shown above on the left) consists of two square plates, each with side length L, separated by a distance d (d < < L). Each plate holds a magnitude of charge Q. Derive expressions for the following in terms of L, d, Q, and physical constants, as appropriate. a. The electric potential difference between the plates. Ans: V = d L 0 b. The capacitance of this capacitor. Ans: C = - c. The electric potential energy stored in the capacitor. Ans: U = ; Qd 2OL The capacitor plates are isolated, so the charge on each plate is held constant. A metal slab (NOT a dielectric) with thickness d/3 is inserted between the plates of the capacitor as shown above on the right. The slab is inserted without touching either plate. The metal slab is at the center of the capacitor and has the same area as the capacitor plates. Derive expressions for the following in terms of L, d, Q, and physical constants. d. The new capacitance of this capacitor. Ans: C 360L2 2d e. The electric potential energy stored in this capacitor. Ans: U = 360L2 The metal slab is removed, allowing the capacitor to return to its original conditions. Now, the plates are connected to a battery, so the potential difference between the plates is held constant (the same potential difference as part (a). The same metal slab is reintroduced into the capacitor at the exact same position in parts (d)-(e). Derive expressions for the following in terms of L, d, Q, and physical constants. f. The new capacitance of this capacitor. Ans: C = 3QL2 2d g. The electric potential energy stored in this capacitor. Ans: U = 30d 460L2 h. The new total charge on the positive plate of the capacitor. Ans: Qnew = 3Q/2