A parallel-plate capacitor has plates of area 0.11 m2 and a separation of 1.1 cm. A battery charges the plates to a potential difference of 130 V and is then disconnected. A dielectric slab of thickness 3.7 mm and dielectric constant 9.0 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 electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab? (a) Number i Units (b) Number i Units (c) Number Units (d) Number Units

A parallel-plate capacitor has plates of area 0.11 m2 and a separation of 1.1 cm. A battery charges the plates to a potential difference of 130 V and is then disconnected. A dielectric slab of thickness 3.7 mm and dielectric constant 9.0 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 electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab? (a) Number i Units (b) Number i Units (c) Number Units (d) Number Units

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

Related questions

Q: 7. (a) The space between two parallel plate capacitor is filled with two dielectric materials of…

Q: A parallel plate capacitor, in which the space between the plates is empty, has a capacitance of Co=…

Q: Three capacitors having capacitances of 6.00, 6.00, and 3.00 µF, respectively, are connected in…

A: Given: Three capacitor connected in series, C1 = 6 μF C2 = 6 μF C3 = 3 μF Voltage =27 V

Q: A capacitor is made from two concentric spheres, one with radius 5.7 cm, the other with radius 7.6…

Q: A certain substance has a dielectric constant of 3.7 and a dielectric strength of 25 MV/m. If it is…

Q: Consider a parallel plate capacitor having plates of area 1.65 cm^2 that are separated by 0.012 mm…

A: C=ε0εrAd=8.854×10-12 C/N·m26.71.65 cm210-4 m21 cm20.012 mm10-3 m1 mm=8.16×10-10 F1 μF10-6…

Q: Consider a parallel plate capacitor having plates of area 1.35 cm2 that are separated by 0.014 mm of…

A: The objective of the question is to calculate the capacitance of a parallel plate capacitor with…

Q: A certain substance has a dielectric constant of 2.8 and a dielectric strength of 18.0 MV/m. If it…

A: Write the given values with the suitable variables. k=2.8E=18 MV/m=18×106 V/mC=9×10-8 FV=4.8…

Q: The two plates of a parallel-plate capacitor each have area1.20 m2, are 4.50 mm apart, and initially…

A: Given, A=1.20 m2d= 4.50 mmV=3.50 kVk=2.50

Q: The figure shows a variable "air gap" capacitor of the type used in manually tuned radios. Alternate…

Q: A parallel plate capacitor has a capacitance of 1.20 nF. There is a charge of magnitude 0.800 μ C on…

Q: parallel plate capacitor has a plate separation d = 1.0mm and a plate area A = 3.0m2. As shown…

Q: A researcher builds a parallel-plate capacitor with adjustable spacing between the plates. When the…

Q: You have two flat metal plates, each of area 1.14 m2, with which to construct a parallel-plate…

A: (a) The equation for the capacitance of a parallel plate capacitor is given by C=ε0Ad Here, ε0 is…

Q: A parallel-plate capacitor, with vacuum in the middle, has a capacitance of 1.2 microFarad. The…

A: Given: Capacitance, C=1.2 μF=1.2×10-6 F Potential, V=12 V Additional charge when a dielectric is…

Q: A 140 pF capacitor is charged to a potential difference of 74 V, and the charging battery is…

Q: A lab technician builds a parallel-plate capacitor with adjustable spacing between the plates. When…

A: Given:The capacitance of the capacitor is The voltage of the battery is .Introduction:The energy…

Q: In the figure the battery has a potential difference of V = 10.0 V and the five capacitors each have…

Q: A certain capacitor has a capacitance of 4μF when its plates are separated by 0.2 mm of vacant…

A: Given: Capacitance, C0=4 μF=4×10-6 F Plate separation, d=0.2 mm=0.2 ×10-3 m Mica sheet thickness,…

Q: A parallel plate capacitor initially charged with a battery that can produce a potential difference…

A: Potential difference = 100VDielectric constant of dielectric material = k= 4.6tickness t=6mm = 0.006…

Q: The plates of a parallel-plate capacitor are charged to a potential difference of 35.0 V. If the…

Concept explainers

Dielectric Constant Of Water

Water constitutes about 70% of earth. Some important distinguishing properties of water are high molar concentration, small dissociation constant and high dielectric constant.

Electrostatic Potential and Capacitance

An electrostatic force is a force caused by stationary electric charges /fields. The electrostatic force is caused by the transfer of electrons in conducting materials. Coulomb’s law determines the amount of force between two stationary, charged particles. The electric force is the force which acts between two stationary charges. It is also called Coulomb force.

Question

the free charge q (c) before and (d) after the slab is inserted? What is the magnitude of the electric field (e) in the spa
plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across
much external work is involved in inserting the slab?
(a) Number
i
Units
(b) Number
i
Units
(c) Number
i
Units
(d) Number
Units
(e) Number
Units
(f) Number
Units
(g) Number
Units
(h) Number
Units

A parallel-plate capacitor has plates of area 0.11 m2 and a separation of 1.1 cm. A battery charges the plates to a potential difference
of 130 V and is then disconnected. A dielectric slab of thickness 3.7 mm and dielectric constant 9.0 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 electric field (e) in the space between the
plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How
much external work is involved in inserting the slab?
(a) Number
i
Units
(b) Number
i
Units
(c) Number
i
Units
(d) Number
Units
(e) Number
Units
(f) Number
Units

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

A certain capacitor has a capacitance of 4μF when its plates are separated by 0.2 mm of vacant space. A battery is used to charge the plate to a potential difference of 500 V and is then disconnected from the system. (a) What will be the potential difference across the plates if a sheet of mica 0.2 mm thick is inserted between the plates? (b) What will the capacitance be after the dielectric is inserted? (c) What is the permittivity of mica?
In a parallel-plate capacitor, the two plates each have an area of 0.460 m2 and are spaced 3.00 mm apart.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, air filled capacitor, has plates of area 0.61 m2 and a separation of 0.1 mm.What is its capacitance? (Units: nF=10-9F)
A capacitor is made with plates of area A = 0.255 m2, separated by a distance d = 0.386 mm. Between the plates is a layer of a material with dielectric constant κ = 2.15. What is the capacitance of the capacitor? Your answer should be in nano-farads (nF)
The capacitance of an empty capacitor is 1.2 μF. The capacitor is connected to a 12-V battery and charged up. With the capacitor connected to the battery, a slab of dielectric material is inserted between the plates. As a result, 2.6 × 10–5 C of additional charge flows from one plate, through the battery, and onto the other plate. What is the dielectric constant of the material?
An air-filled capacitor is made from two flat parallel plates 1.2 mm apart. The inside area of each plate is 8 cm². (a) What is the capacitance of this set of plates? C F (b) If the region between the plates is filled with a material whose dielectric constant is 6.5, what is the new capacitance? C dielectric air = = F