A parallel plate capacitor has a charge of 5.50 × 10¬ C on one plate and -5.50 × 10¬/ C on the other. The distance between the plates is increased by 46.0% while the charge on each plate stays the same. By how much does the energy stored in the capacitor increase? percentage

A parallel plate capacitor has a charge of 5.50 × 10¬ C on one plate and -5.50 × 10¬/ C on the other. The distance between the plates is increased by 46.0% while the charge on each plate stays the same. By how much does the energy stored in the capacitor increase? percentage

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

A parallel-plate capacitor under the key of a keyboard has circular plates that are separated by 0.0500 mm and filled with a dielectric with k = 4.00. When the potential difference across the plates is 5.00 V , the magnitude of the charge on each plate is 8.897 pC. What is the radius of one of the plates? When the key is pressed, the distance between the plates decreases, increasing the capacitance to 1.85 pF. What is the separation between the plates when the key is pressed?
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 255 V. Assume a plate separation of d = 1.55 cm and a plate area of A = 25.0 cm². When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q₁ = 364 PC after Qf = 29133 The charge will be the same before and after the capacitor is submerged in water. Apply the definition of a parallel-plate capacitor, and the relationship among C, Q, and AV. Don't forget to convert from cm to m as needed. pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf= 1.142e-10 F X AV f = 255 This is the potential difference before the capacitor was submerged. When it is in the distilled water, the…
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 260 V. Assume a plate separation of d = 1.29 cm and a plate area of A = 25.0 cm2. When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q; = 445.93 pC after Qf = 445.93 pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = 0.00000000013 AVf = 3.24 (c) Determine the change in energy (in nJ) of the capacitor. AU = -220 Write expressions for the initial and final energies, and take the difference. nJ (d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 260 V potential difference.…
Consider the Earth and a cloud layer 500 m above the planet to be the plates of a parallel-plate capacitor. (a) If the cloud layer has an area of 4.0 km2 = 4000000 m2, what is the capacitance? F (b) If an electric field strength greater than 3.0 ✕ 106 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold? C
Assume you have a charged capacitor that stores 1.10 x 10-12 J of energy while connected to a 1.50-V battery. If this capacitor has circular plates of radius 1.34 cm, how much charge is on these plates when it is fully charged?
Consider the Earth and a cloud layer 8.0 x 10 m above the planet to be the plates of a parallel-plate capacitor. (a) If the cloud layer has an area of 1.0 km² = 1.0 x 10° m2, what is the capacitance? (b) If an electric field strength greater than 3.0 x 10° N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold?
An air-filled parallel-plate capacitor has plates of area 2.40 cm2 separated by 2.50 mm. The capacitor is connected to a 21.0-V battery. (a) Find the value of its capacitance. pF(b) What is the charge on the capacitor? pC(c) What is the magnitude of the uniform electric field between the plates?
The capacitor is charged to a voltage of 4.00 kv using a power source that is then removed. The gap between the plates is then filled by a dielectric layer. The charge on each plate stays constant at 2.50 kv despite the reduction in the potential difference between the plates. Calculate the initial capacitance value of the system.
A parallel-plate capacitor is fully charged by a battery. While still connected to the battery, the distance between the plates increases. What happens to the charge on the capacitor? Answers: decreases increases stays the same
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 225 V. Assume a plate separation of d = 1.47 cm and a plate area of A = 25.0 cm². When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before after AV f (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf= = = (c) Determine the change in energy (in nJ) of the capacitor. AU = nJ AV f (d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 225 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before after…
A parallel plate capacitor with plate separation do is charged by connecting it to a battery. The energy stored by the capacitor is Ug. The battery is then disconnected and the plate separation is changed to d1 = 0.7do. If U1 is the energy stored by the capacitor when the plate separation is di, what is the ratio U1/Uo? O (a) 0.7 O (b) 0.837 O (c) 0.49 O (d) 2.04 O (e) 1.43 O (f) 1.2
Each plate in a parallel-plate capacitor has an area of 0.460 m2 and is separated by 3.00 mm.The capacitor is charged to 4,000 kV using a power source that is subsequently disconnected. The space between the plates is then filled with a layer of dielectric material. Despite the drop in potential difference between the plates, the charge on each plate remains constant at 2.50 kV.Determine the system's initial capacitance value.