A 50 pF parallel-plate capacitor with an air gap between the plates is connected to a...

Question

Question

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 the

science physics

Add a comment Improve this question Transcribed image text

Answer 1

Answer #1

Given that dielectric is introduced between the plates of capacitor when battery is still being connected. Hence after removing the battery, there will be no change in potential difference across the capacitor. Hence New potential difference is 50 V .

Add a comment

Answer 2

Similar Homework Help Questions

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...

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) 250V 50V 25V 100V
A parallel-plate capacitor with a plate area of 50 mm2 and air between the plates can...

A parallel-plate capacitor with a plate area of 50 mm2 and air between the plates can hold 8.0 pC of charge per volt of potential difference across its plates. When a barium titanate dielectric slab completely fills the space between the plates and the capacitor is connected to a 9.0-V battery, what is the electric field magnitude inside the capacitor? IT IS NOT 1.36 * 102 N/C

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 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 plate area 4.0cm^2 and air-gap separation 0.50❝mm is connected to a 9.0-V...

A parallel-plate capacitor with plate area 4.0cm^2 and air-gap separation 0.50❝mm is connected to a 9.0-V battery, and fully charged. The battery is then disconnected. What is the charge on the capacitor? The plates are now pulled to a separation of 0.75?mm. What is the charge on the capacitor now? What is the potential difference between the plates now? How much work was required to pull the plates to their new separation?

A parallel-plate capacitor with plate area 4.60 cm2 and air-gap separation 0.78 mm is connected to...

A parallel-plate capacitor with plate area 4.60 cm2 and air-gap separation 0.78 mm is connected to a 12.00 V battery, and fully charged. The battery is then disconnected. (a) What is the charge on the capacitor? (b) The plates are now pulled to a separation of 0.98 mm. What is the charge on the capacitor now? (c) What is the potential difference across the plates now? (d) How much work was required to pull the plates to their new separation?...
85. A parallel-plate capacitor with plate area 3.0 cm2 and air- gap separation 0.50 mm is...

85. A parallel-plate capacitor with plate area 3.0 cm2 and air- gap separation 0.50 mm is connected to a 12-V battery, and fully charged. The battery is then disconnected. (a) What is the charge on the capacitor? (b) The plates are now pulled to a separation of 0.75 mm. What is the charge on the capacitor now? (c) What is the potential difference between the plates now? (d) How much work was required to pull the plates to their new...
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 only air between the plates is charged by connecting it to a...

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. A. A voltmeter reads 44.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.0 V . What is the dielectric constant of this material? B. What will the voltmeter read if the dielectric is...

A 50 pF parallel-plate capacitor with an air gap between the plates is connected to a...

Homework Answers