Suppose the parallel plates each have an area of 2000 cm? and are 1.00 cm apart. We connect the capacitor to a power supply, charge it to a potential difference V, = 3.00 kV, and disconnect the power supply. We then insert a sheet of nsulating plastic material between the plates, completely filling the space between them. We find that the potential difference decreases to 1.00 kV while the charge on each capacitor plate remains constant. Find the electric field E (a) Vacuum (b) Dielectric Induced charges after the dielectric is inserted.

Suppose the parallel plates each have an area of 2000 cm? and are 1.00 cm apart. We connect the capacitor to a power supply, charge it to a potential difference V, = 3.00 kV, and disconnect the power supply. We then insert a sheet of nsulating plastic material between the plates, completely filling the space between them. We find that the potential difference decreases to 1.00 kV while the charge on each capacitor plate remains constant. Find the electric field E (a) Vacuum (b) Dielectric Induced charges after the dielectric is inserted.

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: AC = 5 mF capacitor is connected to a V = 6V battery and allowed to fully charge. The %3D battery is…

A: Given: Capacitance of parallel plate capacitor C=5mF. The voltage connected is V=6V.

Q: An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm², separated by…

A: Hello. Since your question has multiple sub-parts, we will solve first three sub-parts for you. If…

Q: A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the…

Q: A 68.3-V potential difference is applied across the parallel plates of a 0.15-μF capacitor. (a) What…

Q: A 5.80 µF , parallel-plate, air capacitor has a plate separation of 5.00 mm and is charged to a…

Q: An air-filled parallel-plate capacitor has plates of area 2.10 cm2 separated by 1.80 mm. The…

Q: The potential difference between the plates of a capacitor is 137 V. Midway between the plates, a…

Q: A capacitor consists of two large flat parallel plates. One plate has surface charge density 8.79…

Q: A single, air-filled, capacitor is connected to a battery. The capacitor is fully charged and…

Q: A parallel-plate capacitor with area 0.220 m² and plate separation of 4.20 mm is connected to a…

A: Hi, your question has multiple sub-parts so we will solve first three sub-parts for you. For…

Q: Two parallel-plate capacitors C₁ and C₂ are connected in parallel to a 12.0 V battery. Both…

Q: You are asked to construct a capacitor having a capacitance near 1 nF and a breakdown potential in…

A: Given that, Near capacitance, C = 1nF=. Dielectric constant of…

Q: Imagine that we want to construct a 100 uF parallel-plate capacitor whose metal plates have an area…

Q: An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm?, separated by…

A: Given data *The area of the air-filled capacitor is a = 7.60 cm2 *The given distance between the…

Q: A parallel –plate air capacitor has a capacitance of 350.0 pF and a charge of magnitude 0.500 uC on…

Q: A 5-μF capacitor has 100 V between its terminals. Determine the magnitude of the net charge stored…

A: Given data: Capacitance (C) = 5 μF = 5×10-6 F Voltage (V) = 100 V Required: The magnitude of the…

Q: A parallel-plate air capacitor is to store charge of magnitude 240.0 pC on each plate when the…

Q: The electric field in the region between the plates of a parallel plate capacitor has a magnitude of…

Q: Capacitor with Dielectric N 1. A parallel-plate capacitor has capacitance Co 8.00 pF when there is…

A: The capacitance of the air-filled parallel plate capacitor, C0=8.00pF The distance of separation,…

Q: A parallel-plate capacitor is constructed with circular plates of radius 6.50×10−2 mm . The plates…

Q: An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm, separated by…

A: Given: The area of each parallel plate: A = 7.60 cm2 = 7.60 x 10-4 m2. The distance of separation…

Q: A parallel plate, air filled capacitor, with plate of area A = 7.60 cm^2 and separated by a distance…

A: Write the given values to the suitable variables.Here, A, d, and V are the area of the plate, the…

Q: A parallel-plate capacitor is charged by a 16.0 V battery, then the battery is removed. What is the…

A: If the potential difference between the plates was V0 when the gap between the plate was filled with…

Q: An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by…

A: Given:area of plates A = 7.60 cm2=7.6*10-4 m2distance between plates d=1.4 mm = 1.4*10-3mpotential…

Q: A parallel-plate capacitor with area 0.220 m² and plate separation of 2.00 mm is connected to a…

A: According to the guidelines issued by the company we are supposed to answer only first three sub…

Q: A parallel-plate capacitor under the key of a keyboard has circular plates that are separated by…

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

Suppose the parallel plates each have an area of 2000 cm?
and are 1.00 cm apart. We connect the capacitor to a power
(a) Vacuum
(b) Dielectric
supply, charge it to a potential difference Vo = 3.00 kV, and
disconnect the power supply. We then insert a sheet of
insulating plastic material between the plates, completely
filling the space between them. We find that the potential
difference decreases to 1.00 kV while the charge on each
capacitor plate remains constant. Find the electric field E
+
Induced
charges
after the dielectric is inserted.
100,000 V/m
300,000 V/m
200,000 V/m
1000 V/m

Expert Solution

This question has been solved!

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

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

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

An air-filled parallel-plate capacitor has plates of area 2.10 cm2 separated by 1.80 mm. The capacitor is connected to a(n) 22.0 V battery. (a) Find the value of its capacitance. (b) What is the charge on the capacitor? (c) What is the magnitude of the uniform electric field between the plates?
An air-filled capacitor with C = 4.00 uF is connected to a 6.00 volt battery. While the capacitor remains connected to the battery, a dielectric with K = 2.00 is inserted, completely filling the volume between the plates. What is the energy stored in the capacitor after the dielectric has been inserted? I know the answer is 144 uJ, but am unsure of the step-by-step solution to get this number.
Suppose you want to make a 1.00-F capacitor using two square sheets of aluminum foil. If the sheets of foil are separated by a single piece of paper (thickness of about 0.100 mm and k = 5.00), find the size of the sheets of foil (the length of each edge).
A parallel plate capacitor is charged to a potential of 3000 V and then isolated. Find the magnitude of the charge on the positive plate if the plates area is 0.40 m2 and the distance between the plates is 0.020 m. Group of answer choices 68 µC 54 µC 27 µC 18 µC
A parallel-plate capacitor in air has a plate separation of 1.58 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 270 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. (a) Determine the charge on the plates before and after immersion. before pC pC after (b) Determine the capacitance and potential difference after immersion. Cf = F AV = V (c) Determine the change in energy of the capacitor. nJ Need Help? Read It
An air-filled parallel-plate capacitor has plates of area 2.90 cm? separated by 3.00 mm. The capacitor is connected to a(n) 13.0 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? N/C
Regarding the Earth and a cloud layer 800 m above the Earth as the "plates" of a capacitor, calculate the capacitance. Assume the cloud layer has an area of 1 km2 and that the air between the cloud and the ground is pure and dry. nF Assume charge builds up on the cloud and on the ground until a uniform electric field of 2.0 106 N/C throughout the space between them makes the air break down and conduct electricity as a lightning bolt. What is the maximum charge the cloud can hold? C