A capacitor is made from two flat parallel plates placed 0.40 mm apart. When a charge of 0.020 µC is placed on the plates the potential difference between them is 250 V. a) What is the capacitance? b) What is the area of each plate? c) What is the charge on the plates when the potential difference between them is 500 V? d) What maximum potential difference can be applied between the plates so that the magnitude of electrical fields between the plates does not exceed 3.0 MV/m?

A capacitor is made from two flat parallel plates placed 0.40 mm apart. When a charge of 0.020 µC is placed on the plates the potential difference between them is 250 V. a) What is the capacitance? b) What is the area of each plate? c) What is the charge on the plates when the potential difference between them is 500 V? d) What maximum potential difference can be applied between the plates so that the magnitude of electrical fields between the plates does not exceed 3.0 MV/m?

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: An air-filled parallel-plate capacitor has plates measuring 10.0 cm x 10.0 cm and a plate separation…

A: Given data: Area of first parallel-plate capacitor (A1) = 10.0cm×10.0cm Plate separation for first…

Q: Please answer all four parts to the question.: a) What is the capacitance of this capacitor? b)…

Q: A 2.7-nF parallel plate capacitor with a sheet of Mylar (K = 3.1) filling the space between the…

A: Given That: Capacitance of the capacitor, C0 = 2.7 nF For sheet of Mylar, k = 3.1 Initial potential…

Q: A parallel-plate capacitor is composed of two square metal plates with sides 2 m and separated by…

Q: When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the…

A: Given Data: The potential difference is, V = 150 V. The surface charge density is, σ = 35 nC/cm2.…

Q: 5. A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148 µC on each…

A: Disclaimer: “Since you have asked posted a question with multiple sub-parts, we will solve the first…

Q: If you want to build the equivalent capacitor with a parallel plate capacitor with dielectric…

Q: Two parallel metal plates have a separation of 5.0 cm. If the potential difference between the…

A: Separation, ∆r=5.0 cm=0.05mPotential Difference, ∆V=2000V

Q: IP A parallel-plate capacitor has plates with an area of 1.1x10-2 m² and a separation of 0.82 mm.…

Q: battery connected to a capacitor that originally has a capacitance of 3.0000 Farad (before any…

Q: A parallel plate capacitor has a capacitance of 2.0 nF. The capacitor is initially charged such that…

A: Given, C = 2 nF = 2×10-9 F ∆Vi = 100V k = 5.4

Q: The capacitance of the parallel plate capacitor, whose plates are filled with air and the distance…

A: Given, d = 0.328 mm = 0.328 * 10-3 m C = 245 pF = 245 * 10-12 F Q = 0.148 micro Coulomb We know…

Q: If the area of the plates of a parallel plate capacitor is halved and the separation between the…

Q: When a potential difference of 162 V is applied to the plates of a parallel-plate capacitor, the…

A: C= Q/V =sigmaA/V C is capacitance Q is charge V is the voltage applied Also C= eA/d (e is…

Q: A capacitor has a dielectric with K = 55. Additionally, C = 9 F. The capacitor is charged until the…

Q: Problem 5 Consider a parallel-plate capacitor with a plate area of A = 8.50 cm². The separation…

A: The parallel plate capacitor has a plate area of 8.5 cm2 and a plate separation of 3 mm. The…

Q: The plates of a parallel-plate capacitor are 3.20 mm apart, and each has an area of 9.92 cm2 . Each…

A: Solution: Given Values, Distance(d)=3.20 mm Area(A)=9.92 cm2 Charge(Q)=4.60×10−8 C

Q: A parallel-plate capacitor with plate area A = 0.14 m2 and plate separation d = 0.25 mm is connected…

A: Area , A = 0.14 m2 Separation , d = 0.25 mm = 0.25 x 10-3 m Battery , V0 = 3.5 V Dielectric…

Q: A parallel plate capacitor has a capacitance of 3.707 μF with vacuum between the plates. A…

Q: The plates of a parallel-plate capacitor have a separation of 3.52 mmmm , and each has an area of…

A: (a) The capacitance of the parallel-plate capacitor is

Q: Q3/ A parallel plate capacitor whose plates are (15 cm X 4 cem) and separated by 1 mm air gap and…

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

A capacitor is made from two flat parallel plates placed 0.40 mm apart. When a charge of 0.020 µC is
placed on the plates the potential difference between them is 250 V.
a) What is the capacitance?
b) What is the area of each plate?
c) What is the charge on the plates when the potential difference between them is 500 V?
d) What maximum potential difference can be applied between the plates so that the magnitude of electrical
fields between the plates does not exceed 3.0 MV/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 with 4 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 six UF parallel plate capacitor with a dielectric has a charge of 430 uC on each plate the distance between the plates is .006M and the dielectric constant is 12 determine the area of each plate and the E field between the plates
A parallel plate capacitor is charged up so that the upper plate carries a charge of 0.2C. In a vacuum, the potential difference between the plates is 2V. When a certain dielectric material is placed between the plates, the potential difference drops to 1.5V. Calculate the dielectric constant of the material.
Imagine a parallel plate capacitor with mica between the plates. Suppose the potential difference between the plates is 82V, and there is an electric field between the plates with a strength of 320 N/C. If each plate has an area of 0.030 square meters, a) What is the separation between the plates? b) Find the capacitance of the capacitor. c) How much charge is on each plate? d) How much energy is stored in the capacitor? For each part, explain your reasoning.
Increasing the distance between the plates of an isolated charged capacitor has what effect? A) It increases the electric field between the plates. B) It decreases the energy stored in the capacitor. C) It increases the charge on the plates. D) It increases the voltage difference between the plates
The parallel plates of a capacitor have an area of 0.15 m2 and are separated by 1 mm in air. i) What is the capacitance? ii) If the distance is doubled, what would be the new capacitance? iii) How is the capacitance modified if a dielectric of K=2 is added between the plates?
Area = A dielectric capacitor plates A parallel plate capacitor has each plate area as 3 cm x 4 cm, separated by a 1.0 mm thick cardboard represented by the purple layer in the figure (dielectric constant € = 3.8), is fully charged. Find the capacitance (C) and the maximum charge (Qmax) on the capacitor if it's charged with an electric field E = 16 x 106 V/m.,
A parallel plate capacitor has a potential difference between it's plates of 1.2 V and a plate separation distance of 2.0 mm. What is the magnitude of the electric field if a material that has a dielectric constant of 3.3 is inserted between the plates?
A parallel-plate capacitor filled with air has plates of area 6.5×10−3 m2 and a separation of 0.32 mm. a) Find the magnitude of the charge on each plate when the capacitor is connected to a 11-V battery. (Q = ? nC) b) Why does answer A decrease if the seperation between the plates increases? c) Calculate the magnitude of the charge on the plates if the separation is 0.90 mm. (Q = ? nC)
When a potential difference of 132 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 20.0 nC/cm². What is the spacing between the plates? μm