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What is the maximum charge that can be stored on the 1.90-cm2
plates of an air-filled...
The charge on the 4.00 cm2 area plates of an air-filled parallel plate capacitor is such...
The charge on the 4.00 cm2 area plates of an air-filled parallel plate capacitor is such that the electric field is at the breakdown value. By what factor will the maximum charge on the plates increase when polystyrene is inserted between the plates? (The dielectric strength of air is 3.00 × 106 V/m and that of polystyrene is 2.40 x 107 V/m. The dielectric constant of polystyrene is 2.56. Supporting Materials Physical Constants Additional Materials Tutorial
(a) How much charge can be placed on a capacitor with air between the plates before...
(a) How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 2.00 cm2? (Assume air has a dielectric strength of 3.00 ✕ 106 V/m and dielectric constant of 1.00.) nC (b) Find the maximum charge if polystyrene is used between the plates instead of air. (Assume polystyrene has a dielectric strength of 24.0 ✕ 106 V/m and dielectric constant of 2.56.) nC
The plates of an air-filled parallel-plate capacitor with a plate area of 16.5 cm2 and a...
The plates of an air-filled parallel-plate capacitor with a plate area of 16.5 cm2 and a separation of 8.80 mm are charged to a 130-V potential difference. After the plates are disconnected from the source, a porcelain dielectric with κ = 6.5 is inserted between the plates of the capacitor. (a) What is the charge on the capacitor before and after the dielectric is inserted? Qi = ___C Qf = ____C (b) What is the capacitance of the capacitor after...
A parallel plate capacitor is made of plates of area 0.05 m? each. The plates are...
A parallel plate capacitor is made of plates of area 0.05 m? each. The plates are separated by a distance of 0.200 mm. Initially, the space between the plates is filled with air. (a) What is the capacitance of this air-filled capacitor? (b) If the electric field inside the capacitor exceeds 3.00 x 106 V/m, the air undergoes electrical break- down. (This maximum field is known as the dielectric strength of air.) From this, calculate the maxi- mum voltage (potential...
An air capacitor is made from two flat parallel plates 2.15 mm apart. The magnitude of...
An air capacitor is made from two flat parallel plates 2.15 mm apart. The magnitude of charge on each plate is 0.0153 mu C when the potential difference is 200 V. (a) What is the capacitance? PF (b) What is the area of each plate? m^2 (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of 3.00 times 10^6 V/m.) V (d) When the charge is 0.0153 mu C, what...
An air-filled parallel-plate capacitor has plates of area 2.50 cm2 separated by 3.00 mm. The capacitor...
An air-filled parallel-plate capacitor has plates of area 2.50 cm2 separated by 3.00 mm. The capacitor is connected to a 21.0-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?
1. Two parallel plates, each of area 5.17 cm2, are separated by 4.90 mm. The space...
1. Two parallel plates, each of area 5.17 cm2, are separated by 4.90 mm. The space between the plates is filled with air. A voltage of 4.75 V is applied between the plates. Calculate the magnitude of the electric field between the plates. 2. Calculate the amount of the electric charge stored on each plate. 3. Now distilled water is placed between the plates and the capacitor is charged up again to the same voltage as before. Calculate the magnitude...
The dielectric in a capacitor serves two purposes. It increases the capacitance, compared to an otherwise...
The dielectric in a capacitor serves two purposes. It increases the capacitance, compared to an otherwise identical capacitor with an air gap, and it increases the maximum potential difference the capacitor can support. If the electric field in a material is sufficiently strong, the material will suddenly become able to conduct, creating a spark. The critical field strength, at which breakdown occurs, is 3.0 MV/m for air, but 60 MV/m for Teflon. A parallel-plate capacitor consists of two square plates...
Two parallel plates, each of area 3.37 cm2, are separated by 4.80 mm. The space between...
Two parallel plates, each of area 3.37 cm2, are separated by 4.80 mm. The space between the plates is filled with air. A voltage of 6.25 V is applied between the plates. Calculate the magnitude of the electric field between the plates. Tries 0/20 Calculate the amount of the electric charge stored on each plate. Tries 0/20 Now distilled water is placed between the plates and the capacitor is charged up again to the same voltage as before. Calculate the...
The parallel plates in a capacitor, with a plate area of 9.90 cm2 and an air-filled...
The parallel plates in a capacitor, with a plate area of 9.90 cm2 and an air-filled separation of 2.30 mm, are charged by a 4.10 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 6.50 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.
The charge on the 4.00 cm2 area plates of an air-filled parallel plate capacitor is such that the electric field is at the breakdown value. By what factor will the maximum charge on the plates increase when polystyrene is inserted between the plates? (The dielectric strength of air is 3.00 × 106 V/m and that of polystyrene is 2.40 x 107 V/m. The dielectric constant of polystyrene is 2.56. Supporting Materials Physical Constants Additional Materials Tutorial
A parallel plate capacitor is made of plates of area 0.05 m? each. The plates are separated by a distance of 0.200 mm. Initially, the space between the plates is filled with air. (a) What is the capacitance of this air-filled capacitor? (b) If the electric field inside the capacitor exceeds 3.00 x 106 V/m, the air undergoes electrical break- down. (This maximum field is known as the dielectric strength of air.) From this, calculate the maxi- mum voltage (potential...
An air capacitor is made from two flat parallel plates 2.15 mm apart. The magnitude of charge on each plate is 0.0153 mu C when the potential difference is 200 V. (a) What is the capacitance? PF (b) What is the area of each plate? m^2 (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of 3.00 times 10^6 V/m.) V (d) When the charge is 0.0153 mu C, what...
The parallel plates in a capacitor, with a plate area of 9.90 cm2 and an air-filled separation of 2.30 mm, are charged by a 4.10 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 6.50 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.
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