College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Regarding the Earth and a cloud layer 800 m above the Earth as the "plates" of a capacitor, calculate the capacitance. Assume the cloud layer has an area of 1 km2 and that the air between the cloud and the ground is pure and dry.
nF
Assume charge builds up on the cloud and on the ground until a uniform electric field of 2.0 106 N/C throughout the space between them makes the air break down and
C
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- An air-filled parallel-plate capacitor has plates of area 2.40 cm2 separated by 2.50 mm. The capacitor is connected to a 21.0-V battery. (a) Find the value of its capacitance. pF(b) What is the charge on the capacitor? pC(c) What is the magnitude of the uniform electric field between the plates?arrow_forwardA parallel-plate capacitor has capacitance C0 = 8.00 pF when there is air between the plates. The separation between the plates is 1.20 mm. Express your answer with the appropriate units. Part A: What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00×104 V/m? Part B: A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00×104 V/m?arrow_forwardAn air-filled parallel-plate capacitor has plates of area 2.80 cm separated by 1.80 mm. The capacitor is connected to a(n) 18.0 v battery. (a) Find the value of its capacitance. pF (b) What is the charge on the capacitor? pC (c) What is the magnitude of the uniform electric field between the plates? N/Carrow_forward
- Consider a natural parallel plate capacitor where one plate is given by some clouds in the sky and and the other plate is the surface of the earth directly under the cloud. We have a situation where the clouds are 740m above the ground.a) If we consider a cloud area of 1.2km2 what is the capacitance of this capacitor? b) Suppose that the capacitor discharges when the electric field is larger than 5.7 × 105 N/C (lightning!). What is themaximum charge the cloud can hold?(E0 = 8.85 × 10−12 C2/Nm2)arrow_forwardRegarding 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/marrow_forwardA parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.30 and whose dielectric strength is 1.20 x 108 V/m. The desired capacitance is 0.300 μF, and the capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates. m²arrow_forward
- Show transcribed image text An electric field is given by Ex = 3.5x^3 kN/C. Find the potential difference between the points on the x axis at x = 1 m and x = 4 m. kV Find the maximum surface charge density sigma max that can exist on the surface of any conductor before dielectric breakdown of a gas with a dielectric strength of 3.95 MV/m occurs. C/m^2arrow_forwardAn air-filled parallel-plate capacitor has plates of area 2.50 cm^2 separated by 1.80 mm. The capacitor is connected to a(n) 13.0 V battery. a) Find the value of its capacitance. pF b) What is the charge on the capacitor? pC c) What is the magnitude of the uniform electric field between the plates? N/Carrow_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
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