A parallel-plate capacitor with air between its plates is connected to an 80.0 V battery. When...
A parallel-plate capacitor with air between its plates is connected to an 80.0 V battery. When fully charged, the capacitor has an energy of 130 nJ. Without disconnecting the battery, a slab of dielectric is inserted between the plates of the capacitor, fully filling the gap. The energy stored in the capacitor is now 410 nJ. The area of each plate of the capacitor is 0.005 m2.
(e) What is the electric field in the air-filled capacitor?
(f) What is the electric field in the dielectric-filled capacitor?
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A parallel-plate capacitor with air between its plates
is connected to an 80.0 V battery. When...
A parallel-plate capacitor with air between its plates is connected to an 80.0 V battery. When...
A parallel-plate capacitor with air between its plates is connected to an 80.0 V battery. When fully charged, the capacitor has an energy of 130 nJ. Without disconnecting the battery, a slab of dielectric is inserted between the plates of the capacitor, fully filling the gap. The energy stored in the capacitor is now 410 nJ. (c) What is the dielectric constant of the dielectric? (d) What is the charge on the dielectric-filled capacitor?
A parallel-plate capacitor with air between its plates is connected to an 80.0 V battery. When...
A parallel-plate capacitor with air between its plates is connected to an 80.0 V battery. When fully charged, the capacitor has an energy of 130 nJ. (a) What is the capacitance of the capacitor? (b) What is the charge on the capacitor (i.e., the charge on its positive plate)?
A 2.0 μF parallel-plate air-filled capacitor is connected across a 10 V battery. (a) Determine the...
A 2.0 μF parallel-plate air-filled capacitor is connected across a 10 V battery. (a) Determine the charge on the capacitor and the energy stored in the capacitor. (b) An identical 2.0 μF parallel-plate air-filled capacitor is connected across a 5 V battery, and a dielectric slab with dielectric constant κ is inserted between the plates of the capacitor, completely filling the region between the plates, while the battery remains connected. The energy stored in this capacitor is four times that...
A parallel-plate capacitor filled with air carries a charge Q. The battery is disconnected, and a...
A parallel-plate capacitor filled with air carries a charge Q. The battery is disconnected, and a slab of material with dielectric constant k = 2 is inserted between the plates. Which of the following statements is correct? The voltage across the capacitor decreases by a factor of 2. The voltage across the capacitor is doubled. The electric field is doubled. The charge on the plates decreases by a factor of 2. The charge on the plates is doubled.
Two air-filled parallel-plate capacitors with capacitances C1 and C2 are connected in series to a battery...
Two air-filled parallel-plate capacitors with capacitances C1 and C2 are connected in series to a battery that has voltage V; C1 = 3.00 μF and C2 = 6.00 μF. The electric field between the plates of capacitor C2 is E02. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K = 4 is inserted between the plates of capacitor C1, completely filling the space between them. After the dielectric is inserted in C1, the electric...
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 50 pF parallel-plate capacitor with an air gap between the plates is connected to a...
A 50 pF parallel-plate capacitor with an air gap between the plates is connected to a 50V battery. A Teflon (K=2) slab is then inserted between the plates and completely fills the gap. Then the battery is disconnected. What is the new potential difference? (1 pF = 10-12 F)
The figure shows a parallel-plate capacitor of plate area A and plate separation d. A potential differenceV0 is applied between the plates.
The figure shows a parallel-plate capacitor of plate area A and plate separation d. A potential differenceV0 is applied between the plates. While the
battery remains connected, a dielectric slab of thickness b and dielectric constant κ is placed between the plates
as shown. Assume A = 130 cm2, d = 1.94
cm, V0 = 72.6 V, b = 0.735 cm, and κ =
3.15. Calculate (a) the capacitance,(b) the charge on the capacitor plates,(c) the electric field in the gap, and(d)...
A 193 volt battery is connected to a 0.3 microfarad parallel-plate capacitor with air between the...
A 193 volt battery is connected to a 0.3 microfarad parallel-plate capacitor with air between the plates. Now the battery is disconnected, with care taken not to discharge the capacitor. Some paper is then inserted between the plates completely filling up the space. What is the final potential difference between the plates ? (the dielectric constant for air is = 1 and for the paper inserted is = 3)
A 50 pF parallel-plate capacitor with an air gap between the plates is connected to a...
A 50 pF parallel-plate capacitor with an air gap between the plates is connected to a 50V battery. A Teflon (K=2) slab is then inserted between the plates and completely fills the gap. Then the battery is disconnected. What is the new potential difference? (1 pF - 10-12 F) 50V 100V 25V 250V
A 50 pF parallel-plate capacitor with an air gap between the plates is connected to a 50V battery. A Teflon (K=2) slab is then inserted between the plates and completely fills the gap. Then the battery is disconnected. What is the new potential difference? (1 pF = 10-12 F)
A 50 pF parallel-plate capacitor with an air gap between the plates is connected to a 50V battery. A Teflon (K=2) slab is then inserted between the plates and completely fills the gap. Then the battery is disconnected. What is the new potential difference? (1 pF - 10-12 F) 50V 100V 25V 250V
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