•49 Figure 25-48 shows a parallel-plate ca-pacitor with a plate area A = 7.89 cm2 andplate separation d = 4.62 mm. The top half ofthe gap is filled with material of dielectricconstant K = 11.0; the bottom half is filledwith material of dielectric constant K2 = 12.0.What is the capacitance?K2Figure 25-48Problem 49.

Given : area of plate, A = 7.89 cm2 = 7.89 × 10-4 m2 separation between the plates, d =…

Answered: •49 Figure 25-48 shows a parallel-plate…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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5- None of the above (7) A capacitor has circular plates of radius 10 cm separated by 1 mm. Its capacitance is (a)- 1- 8.85 x 102F (b)- 2.78 x 10" F (c)- 1.11 x 10° F (d) 2.78 x 101° F 5- None of the above
#5
95.)A sheet of insulating plastic material is inserted between the plates without otherwise disturbing the system What effect does this have on the capacitance? (A) It causes the capacitance to increase. (B) It causes the capacitance to decrease. (C) None; the capacitance does not change. (D) Nothing can be said about the effect without knowing the dielectric constant of the plastic. (E) Nothing can be said about the effect without knowing the thickness of the sheet.
I. Determine the unknown quantities on each circuit. 1.) C5 μF 2.) C1 = 1 µF C3 = 2 µF Find: Find: C2 = 2 µF C3 = 3 µF а. Ст а. Ст b. qт C,2 μF b. qT c. q1, q2, q3 d. V1, V2, V3 c. 91, q2, q3, q4, q5 C4 = 3 µF C5 = 1 µF d. V1, V2, V3, ,V4, V5 20 V 20 V II. Determine the total capacitance on each circuit. 1.) 2.) 9 μF 12 μF 8 µF + 4 µF 20 µF 12 μF H 5 µF 3 µF 5 µF 3.) 3 µF 4.) 3 µF 5.) 7 µF 5 µF 3 µF HAE H 5 μF 4 µF HH 2 µF
2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K1 3 and K, = 5, as shown in Figure. Parallel-plates have the area A = 0.7 m². The distance is d = 0.05 m. A potential difference of AV = 24 V is applied to the circuit. (a) Find the equivalent capacitance of the system. (b) Find the potential AV1. (8=8.85×10-12 C²/N×m²) K1 A/2 AV1 K2 A/2 K2 K1 A/2 2d/3 d/3 K2 A/2 AV = 24 V
12) Please help on the following question!
What could be an option to vary the capacitance of a capacitor even though the plate areas and supply voltage are kept constant? by tilting one plate with reference to the other by changing the plate separation by breaking one plate by sliding one plate with reference to the other
Could some one explain this thanks
12 : Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. 'V'. If one of the capacitor is completely filled with dielectric material of constant 'K', then potential difference of the other capacitor will become KV (a) V(K + 1) (b) K +1 К -1 V (c) KV (d) Κ(Κ + 1)
Solve problem in image thank u ASAP

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