concept used to solve part C is that on disconnecting a batterfrom the charged capacitor, charge over it remains same while if battery remains cconnected, potential remains same
A parallel plate capacitor is comprised of two metal plates with area A and separated by...
An air-filled parallel-plate capacitor has plate area A and plate separation d. The capacitor is connected to a battery that creates a constant voltage V.A) Find the energy U_0 stored in the capacitor. Express your answer in terms of A, d, V, and ϵ_0.B) The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separation reaches 3d. Find the new energy U_1 of the capacitor after this process. Express...
explain why please A parallel-plate capacitor is made of two conducting plates of area A separated by a distance d. The capacitor carries a charge Q and is initially connected to a battery that maintains a constant potential difference between the plates. The battery is then disconnected from the plates and the separation between the plates is doubled. ) Which of the following remains constant? Voltage across the capacitor Capacitance of the capacitor Charge on the capacitor Submit (Survey Question)...
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...
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 parallel plates in a capacitor, with a plate area of 9.00 cm2 and an air-filled separation of 3.30 mm, are charged by a 5.40 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.10 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 parallel plates in a capacitor, with a plate area of 9.00 cm2 and an air-filled separation of 3.30 mm, are charged by a 5.40 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.10 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 parallel plates in a capacitor, with a plate area of 5.30 cm2 and an air-filled separation of 4.60 mm, are charged by a 3.60 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 6.00 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.
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...
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)...
2) A 9 volt battery is connected to a parallel plate capacitor with an initial capacitance of 5 micro-farads without a dielectric. While still connected to the battery, the sheets are moved a factor of 6 times further apart and a dielectric with a dielectric constant of 1.8 is inserted between the sheets. Then, the battery is disconnected, and after the battery is disconnected, the dielectric is removed and the sheets are brought a factor of 4 times closer together....