Question

A parallel-plate capacitor is charged using a 100 V battery, then the battery is removed. If...

A parallel-plate capacitor is charged using a 100 V battery, then the battery is removed. If a dielectric slab is slid between the plates, filling the space inside, the capacitor voltage drops to 30 V. What is the electric constant of the dielectric?

0 0
Add a comment Improve this question Transcribed image text
Know the answer?
Add Answer to:
A parallel-plate capacitor is charged using a 100 V battery, then the battery is removed. If...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 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...

  • 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

  • A parallel-plate vacuum capacitor is connected to a batteryand charged until the stored electric energy is...

    A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is U. The battery is removed, and then a dielectric material with dielectric constant K is inserted into the capacitor, filling the space between the plates. Finally, the capacitor is fully discharged through a resistor (which is connected across the capacitor terminals).A.)Find Ur, the the energy dissipated in the resistor.Express your answer in terms of U and other given quantities.B.) Consider the same situation...

  • 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 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.)?...

  • 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 stays connected. 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: O stays the same O increases twice O decreases twice O decreases 4 times O increases 4 times

  • Constants Part A A parallel-plate capacitor with only air between the plates is charged by connecting...

    Constants Part 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 voltmeter reads 49 0 V when placed across the capacitor When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 14.5 V What is the dielectric constant of this material? Part B What will the voltmeter read il...

  • 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 parallel plate capacitor of 2 pF is charged to 12 V and then isolated from...

    A parallel plate capacitor of 2 pF is charged to 12 V and then isolated from the power source. The space between the plates is now filled with a dielectric plate sheet whose dielectric constant is 5.0. The new voltage across the capacitor is with steps please

  • A parallel plate capacitor with plate separation of 1 millimeter is connected to a battery of...

    A parallel plate capacitor with plate separation of 1 millimeter is connected to a battery of 10 volts. If the region between the plates is free space (epsilon_r = 1), calculate the electric field between the plates in Volts/meter, and the electric flux density between the plates in Coulombs/m^2. Suppose a dielectric with dielectric constant epsilon_r = 5 is inserted between the plates. With the battery still connected, calculate the electric field and the electric flux density. Which field has...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT