College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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An air-filled parallel-plate capacitor has plates of area 2.30 cm2 separated by 1.50 mm. The capacitor is connected to a 12.0-V battery. (a) Find the value of its capacitance. (b) What is the charge on the capacitor? (c) What is the magnitude of the uniform electric field between the plates?
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- A shark is able to detect the presence of electric fields as small as 1.70 μV/m. To get an idea of the magnitude of this field, suppose you have a parallel plate capacitor connected to a 2.00-V battery. How far apart must the parallel plates be to have an electric field of 1.70 μV/m between the plates? need answer in kmarrow_forwardTwo parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.30×105 V/m. When the space is filled with dielectric, the electric field is E= 2.50×105 V/m. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Energy density, both before and after the. What is the charge density on each surface of the dielectric? Express your answer in coulombs per meter squared. να ΑΣΦ ? |X| X•10" |oi| = C/m²arrow_forwardA capacitor is composed of two metal plates. The two plates have the dimensions L = 0.11 m and W = 0.56 m. The plates have a distance between them of d = 0.1 m, and are parallel to each other. Part (a) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an expression for the magnitude of the electric field, |E|, halfway between the plates. Part (b) Input an expression for the magnitude of the electric field, |E2|, just in front of plate two. Part (c) If plate two has a total charge of q = -1 mC, what is its charge density, σ, in C/m2?arrow_forward
- A 1.8 mm by 2.6 mm plate capacitor has the plates separated by a distance of 0.12 mm. (a) When a charge 4.00 × 10 −11 C of charge is placed on the capacitor, what is the electric field between the plates? (b) If the electric field with the dielectric constant 130 is placed between the plates while the charges on the capacitor stays the same, what is the dielectric field in the dielectricarrow_forwardA slab of copper of thickness b = 1.68 mm is thrust into a parallel-plate capacitor of plate area A = 1.96 cm2 and plate separation d = 5.35 mm, as shown in the figure; the slab is exactly halfway between the plates. (a) What is the capacitance after the slab is introduced? (b) If a charge q = 2.68 µC is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? (c) How much work is done on the slab as it is inserted? (d) Is the slab sucked in or must it be pushed in? Copper (a) Number 4.73e-13 Units (b) Number i 0.686 (c) Number i 5.36e-10 Units J (d) sucked inarrow_forwardConsider a parallel-plate capacitor having an area of 2550 mm^2 and a plate separation of 4.9mm and with a material of dielectric constant 5.9 positioned between the plates. Also, the value of E0 is 8.85x10^-12F/m. a) What is the capacitance of this capacitor in pF? b) compute the electric field that must be applied for a charge pf 7.8x10^-8C to be stored on each plate in V/m.arrow_forward
- A 5.30 cm by 2.40 cm parallel plate capacitor has the plates separated by a distance of 2.00 mm. When 4.00 × 10−11 C of charge is placed on this capacitor, what is the electric field between the plates?arrow_forwardThe energy density in the region between the plates of a capacitor is 0.01 J/m^3. What is the magnitude of the electric fields between the plates? (E0=8.854 x 10^-12 C^2/Nm^2)arrow_forward
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