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4. E = V / d = (3000 V) / 0.02 m
E = 1.5 x 10^5 V/m
5. C = e0 A / d = (8.854 x 10^-12)(0.40)/0.02
C = 1.8 x 10^-10 F
6. Q = C V = (1.8 x 10^-10)(3000) = 5.4 x 10^-7 C
7. W = q delta(V) = (-4 x 10^-6)(0 - 3000)
W = 1.2 x 10^-3 J
8. Q = C V = constant
C = (1.8 x 10^-10)(3000)/1000
C = 5.4 x 10^-10 F
9. k = Cf/ Ci = 5.4/1.8 = 3
Problems 4-9 refer to this same capacitor: The plates of a parallel plate capacitor each have...
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A parallel-plate capacitor has a plate area of A = 250 cm2 and a separation of d = 2.00 mm. The capacitor is charged to a potential difference of V0 = 150 V by a battery. A dielectric sheet (κ = 3.50) of the same area but thickness ℓ = 1.00 mm is placed between the plates without disconnecting the battery. (See figure 24-18 on page 642). Determine the initial capacitance of the air-filled capacitor. Determine the charge on the...
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Suppose the parallel plates in Fig. 24.15 each have an area of 2000 cm2 (2.00 x 101m2) and are 1.00 cm (1.00 x 104m) apart, we connect the capacitor to a power supply, charge it to a potential difference V。. 3.00kv, and disconnect the power supply. We then insert a sheet of insulating plastic material between the plates, completely filling the space between them. We find that the potential difference decreases to 1.00kV while the charge on each capacitor plate...
A parallel-plate capacitor has capacitance Cq = 7.00 pF when there is air between the plates. The separation between the plates is 1.10 mm. What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 x 10^4 V/m? A dielectric with K = 2.60 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what...
A parallel-plate capacitor has a 4 mm plate separation, 0.5 m^2 surface area per plate, and a dielectric with E_r = 6.8. If the plates are maintained at 9 V potential difference, calculate the capacitance, the charge density on each plate.
A parallel-plate capacitor has a plate area of A = 250 cm2 and a separation of d = 2.00 mm. The capacitor is charged to a potential difference of V0 = 150 V by a battery. A dielectric sheet (κ = 3.50) of the same area but thickness ℓ = 1.00 mm is placed between the plates without disconnecting the battery. (See figure 24-18 on page 642). Determine the electric field in the dielectric. Determine the free charge on the...
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A parallel-plate capacitor (Homework #2) A parallel-plate capacitor is made by a pair of plates with area A and separation d. At first, the plates are charged up to +Q and -O respectively, and then a battery is disconnected. Express all the answers for a) through f) only by Q, d, A and 5 +Q 2d -Q Area A a) What is the electric field E inside of the capacitor? b) What is the potential difference V between the plates?...