Homework Answers
Since capacitor is still connected to the battery, while dielectric slab is inserted, battery will maintain the voltage and therefore electric field to remain same between the plates.
Answer: Stays the same
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Question 4 1 pts A parallel-plate capacitor of capacitance C has the area of plates A...
Question 4 1 pts A parallel-plate capacitor of capacitance C has the area of plates A...
Question 4 1 pts A parallel-plate capacitor of capacitance C has the area of plates A and distance between plates d. Capacitor is connected to the battery of voltage V, fully charged and then disconnected. A slab of dielectric material with dielectric constant 4 is then inserted into capacitor, completely filling region between plates. Electric field between plates of capacitor: decreases twice stays the same increases twice decreases 4 times increases 4 times
Question 2 1 pts A parallel -plate capacitor of capacitance C with distance d between plates...
Question 2 1 pts A parallel -plate capacitor of capacitance C with distance d between plates is fully charged from a battery of voltage V and stays connected. The plates of capacitor are slowly pulled to the half of initial distance (d/2). Energy stored is capacitor after pulling the plates to distance d/2: Increased 4 times Increased twice Decreased twice Decreased 4 times Stays the same
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...
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 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 of capacitance Co, plate area A, spacing d is charged to voltage V....
A parallel-plate capacitor of capacitance Co, plate area A, spacing d is charged to voltage V. and then disconnected from the charging battery. A slab with dielectric constant K and thickness d/2 is thrust into the capacitor, as shown in the figure below; the slab is exactly halfway between the plates. к (a) What is the new capacitance in terms of Co? (b) What is the ratio of the stored energy before to that after the slab is inserted (U/0.)?...
Which of the following would increase the capacitance of a parallel-plate capacitor? I. Insert a dielectric...
Which of the following would increase the capacitance of a parallel-plate capacitor? I. Insert a dielectric between the plates. II. Increase the surface area of each plate. III. Increase the separation distance between the plates. O I and II only OII and III only All of the above. A capacitor is charged with a battery to a voltage V and then disconnected from the battery. A dielectric is inserted between the plates. When the dielectric is inserted, what happens to...
A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the following...
A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the following quantities increase, decrease, or remain the same as equilibrium is reestablished? 1. Charge on plates (plates remain connected to battery) 2. Electric potential energy (plates isolated from battery before inserting dielectric) 3.Capacitance (plates isolated from battery before inserting dielectric) 4. Voltage between plates (plates remain connected to battery) 5. Charge on plates (plates isolated from battery before inserting dielectric) 6. Capacitance (plates remain connected...
A parallel-plate capacitor has plate area of 0.12 m2and plate separation 1.2 cm. It is charged...
A parallel-plate capacitor has plate area of 0.12 m2and plate separation 1.2 cm. It is charged by a battery to potential difference of 120 V, then disconnected. A dielectric slab, thickness 4.0 mm and dielectric constant 4.8, is placed symmetrically between the plates.(a) Calculate the capacitance before and after the slab is inserted.(b) Calculate the free charge q before and after the slab is inserted.(c) Calculate the magnitude of the electric field in the space between the plates and the...
A parallel-plate capacitor has plates of area 0.13 m2 and a separation of 1.9 cm. A battery...
A parallel-plate capacitor has plates of area 0.13 m2 and a separation of 1.9 cm. A battery charges the plates to a potential difference of 150 V and is then disconnected. A dielectric slab of thickness 9.3 mm and dielectric constant 4.0 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? What is the free charge q (c) before and (d) after the slab is inserted? What is the magnitude of the...
Question 4 1 pts A parallel-plate capacitor of capacitance C has the area of plates A and distance between plates d. Capacitor is connected to the battery of voltage V, fully charged and then disconnected. A slab of dielectric material with dielectric constant 4 is then inserted into capacitor, completely filling region between plates. Electric field between plates of capacitor: decreases twice stays the same increases twice decreases 4 times increases 4 times
Question 2 1 pts A parallel -plate capacitor of capacitance C with distance d between plates is fully charged from a battery of voltage V and stays connected. The plates of capacitor are slowly pulled to the half of initial distance (d/2). Energy stored is capacitor after pulling the plates to distance d/2: Increased 4 times Increased twice Decreased twice Decreased 4 times Stays the same
A parallel-plate capacitor of capacitance Co, plate area A, spacing d is charged to voltage V. and then disconnected from the charging battery. A slab with dielectric constant K and thickness d/2 is thrust into the capacitor, as shown in the figure below; the slab is exactly halfway between the plates. к (a) What is the new capacitance in terms of Co? (b) What is the ratio of the stored energy before to that after the slab is inserted (U/0.)?...
Which of the following would increase the capacitance of a parallel-plate capacitor? I. Insert a dielectric between the plates. II. Increase the surface area of each plate. III. Increase the separation distance between the plates. O I and II only OII and III only All of the above. A capacitor is charged with a battery to a voltage V and then disconnected from the battery. A dielectric is inserted between the plates. When the dielectric is inserted, what happens to...
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