An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.90 mm. If a 20.8-V potential difference is applied to these plates, calculate the following. (a) the electric field between the plates - magnitude & direction (b) the capacitance
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Q: An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by…
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Q: An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by…
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Q: An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm, separated by…
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Q: An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm², separated by…
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- A parallel-plate capacitor with area 0.220 m² and plate separation of 2.00 mm is connected to a 7.40-V battery. (a) What is the capacitance? F (b) How much charge is stored on the plates? C (c) What is the electric field between the plates? N/C (d) Find the magnitude of the charge density on each plate. C/m2 (e) Without disconnecting the battery, the plates are moved farther apart. Qualitatively, what happens to each of the previous answers?Consider a parallel-plate capacitor whose (circular) plates are separated by a 3.0-mm-thick slab of glass with dielectric constant 5.0. The diameter of each plate is 4.0 cm, and the slab fills the entire volume between the plates. The electric field between the plates with the dielectric in place has magnitude 8.0 x 103 V/m. (a) Find the magnitude of the charge on either plate of the capacitor. (b) Now suppose the slab of glass is removed, leaving an air gap between the plates. Calculate the total energy stored in this capacitor and the energy density (energy per unit volume).A certain parallel-plate capacitor consists of two plates, each with area 2.00 x 102cm2, and separated by a gap of 0.400 cm. (Assume air as the dielectric material.) a) What is the capacitance of this capacitor? b) If the capacitor is connected across a 500.0 V source, what charge will it hold? c) What is the magnitude of the electric field between the plates of the capacitor? (under the conditions of part b)?
- An air-filled parallel-plate capacitor has plates of area 2.10 cm2 separated by 1.80 mm. The capacitor is connected to a(n) 22.0 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?A parallel-plate capacitor has circular plates of 11.3 cm radius and 1.17 mm separation. (a) Calculate the capacitance. (b) What charge will appear on the plates if a potential difference of 129 V is applied? (a) Number i (b) Number i Units UnitsAn air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm², separated by a distance of 1.50 mm. A 25.0- V potential difference is applied to these plates. (a) Calculate the electric field between the plates. |kV/m (b) Calculate the surface charge density. nC/m² (c) Calculate the capacitance. pF (d) Calculate the charge on each plate. pC
- Regarding the Earth and a cloud layer 650 m above the Earth as the "plates" of a capacitor, calculate the capacitance if the cloud layer has an area of 1.53 km2. If an electric field of 2.0 × 106 N/C makes the air breakdown and conduct electricity (lightning), what is the maximum charge the cloud can hold? εo = 8.85 x 10−12 F/mA 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 inThe electric field in a region of space has the components Ey = E₂ = 0 and Ex = (5.20 N/(C-m))x. Point A is on the y axis at y = 4.70 m, and point B is on the x axis at x = 5.30 m. What is the potential difference (in V) VB - VA?