A parallel plate capacitor has area 15 square centimeters and separation distance 2 millimeters. If the charge stored on the positive plate is 3 E-12 C, calculate the magnitude of the electric field between the plates. 2.3 N/C 230 N/C 2 E-9 N/C 0.45 N/C 23000 N/C

A parallel plate capacitor has area 15 square centimeters and separation distance 2 millimeters. If the charge stored on the positive plate is 3 E-12 C, calculate the magnitude of the electric field between the plates. 2.3 N/C 230 N/C 2 E-9 N/C 0.45 N/C 23000 N/C

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: Two 2.2-cm-diameter-disks spaced 1.7 mm apart form a parallel-plate capacitor. The electric field…

Q: A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.153 μC on…

A: in this question we use simple equation of capacitance.

Q: An air filled capacitor is connected to a 3.0V battery. Each plate of the capacitor has area of…

A: Voltage between plates (V) = 3.0 VArea (A) = 0.7 m2Seperation (d) = 1.5 mm = 1.5×10-3 mRequired:The…

Q: A parallel-plate capacitor is formed from two 6.0cm diameter electrodes spaced2.0mm apart. The…

Q: A parallel-plate air capacitor with a capacitance of 250 pF has a charge of magnitude 0.146 μC on…

Q: Two parallel plates have equal and opposite charges. When the space between the plates is evacuated,…

A: Given,The electric field between the plates, E= 5.40×105 V/mIf the space between the space is filled…

Q: A parallel plate capacitor has a plate area of 21.4 x 10-4 m2 and a plate separation distance of 27…

Q: A gold(III) ion is accelerated by the electric field created between two parallel plates separated…

A: As charge is placed in an electric field due to electric force there will be work done on the…

Q: A parallel-plate capacitor is formed from two 2.0 cmcm ××2.0 cmcm electrodes spaced 2.6 mmmm apart.…

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

A: Given: The area of the plates is 1.20 m2. The distance between the plates is 4.50 mm. The power…

Q: A capacitor is created by two metal plates. Each plate has dimensions L = 0.39 m and W = 0.38 m. The…

A: The objective of the question is to find the expressions for the magnitude of the electric field…

Q: The uniform electric field between two parallel plates of opposite charges and separated by 40 mm…

A: Given Electric field (E) = 2 x 10³ N/C Separation between the plates d = 40 mm Charge on the…

Q: The electric field strength is 15500.0N/C inside a parallel plate capacitor with a 1.24mm spacing.…

Q: In A certain parallel-plate capacitor is filled with a dielectric for which k = 5.14. The area of…

Q: An electron is placed very close to the negative plate of a parallel plate capacitor. This electron…

Q: A 5.30 cm by 2.40 cm parallel plate capacitor has the plates separated by a distance of 2.00 mm.…

A: The objective of the question is to find the electric field in the dielectric of a parallel plate…

Q: plate of a parallel-plate capacitor is a square of side 0.0449 m, and the plates are separated by…

Q: A parallel-plate capacitor consists of two identical rectangular plates each with a length of 4.0 cm…

A: Given data: A parallel-plate capacitor Length (L) = 4.0 cm = 4.0×10-2 m Width (b) = 1.0 cm =…

Q: The uniform electric field between two parallel plates of opposite and separate loads of 40 mm has…

A: Positive charge flows in the direction of electric field while negative charge flows opposite to the…

Q: An air-filled parallel-plate capacitor has plates of area 2.70 cm2 separated by 1.00 mm. The…

Q: Oppositely charged parallel plates are separated by 4.25 mm. A potential difference of 600 V exists…

Q: Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE,…

A: The expression for the charge on a capacitor is,

Q: A parallel plate capacitor is formed of two 10 cm × 10 cm plates spaced 1 cm apart. The plates are…

A: Given Charge on positive plate is 1nC Charge on negative plate is -1nC. Q = 1.67*10-19C m =…

Q: Each plate of a parallel-plate capacitor is circular and has diameter 24.0 cm. The plates are 15.0…

A: diameter = 24 cm distance between plates (d) = 15 μmCharge (Q) = ± 8μC

Q: A charged capacitor consisting of two parallel plates has an area and separate of 0.500 m2 and 9.00…

Q: A parallel-plate capacitor is formed from two 9.0 cm × 9.0 cm electrodes spaced 3.0 mm apart. The…

Q: A 4 F parallel-plate capacitor with a dielectric has a charge of 42 μC on each plate. The distance…

A: Solution:- Given- Capacitance(c)= 4μF charge (Q)= 42μc distance(d)= 0.007m…

Q: A parallel plate capacitor has capacitance C0 = 15.0 pF when there is air between the plates. The…

Q: A shark is able to detect the presence of electric fields as small as 1.70 μV/m. To get an idea of…

A: Electric field (E) = 1.70 μVmVoltage of battery (V) = 2 volts

Q: Air "breaks down" when the electric field strength reaches 3××1066 N/CN/C, causing a spark. A…

A: *The given electric field is E = 3 × 106 N/C*The plate area is A = (4.5 cm × 4.5 cm)…

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 parallel plate capacitor has area 15 square centimeters and separation distance 2 millimeters. If the charge stored on the positive plate is 3 E-12 C, calculate the magnitude of
the electric field between the plates.
2.3 N/C
230 N/C
2 E-9 N/C
0.45 N/C
23000 N/C

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

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 capacitor is composed of two metal plates. The two plates have the dimensions L = 0.11 m and W = 0.56 m. The plates have a distance between them of d = 0.1 m, and are parallel to each other. Part (a) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an expression for the magnitude of the electric field, |E|, halfway between the plates. Part (b) Input an expression for the magnitude of the electric field, |E2|, just in front of plate two. Part (c) If plate two has a total charge of q = -1 mC, what is its charge density, σ, in C/m2?
a 8 μf parallel plate capacitor with a dielectric has a charge of 543 μc on each plate. the distance between the plates is 0.008m and the dielectric constant is 22. determine the area of each plate and the e field between the plates.
the electric field of 2 plates is E=σ/2ε0 find V (electric potential) by integrating the electric field
A charged air-filled parallel plate capacitor of 22 nF has a energy density of 30x10-3 J/m³. If the distance between the plates is 200 µm, calculate the magnitude of the charge, in µC, of the capacitor.
Consider a parallel-plate capacitor with plate separation d, plate area A, whose plates have charge ±Q. A particle of charge q < 0 and mass m is released from rest at the negative plate of the capacitor and allowed to accelerate towards the positive plate. With what speed does the charge strike the positive plate? Answer in terms of d, A, Q, q, m, and/or e0.
The capacitance of a parallel plate capacitor was measured as 36 pF. Since the surface area of the plates is 2x10-4 m2 and the distance between the plates is 2x10 -4 m, calculate the dielectric constant of the insulator placed between the plates? (E0=9x10-12 F/m)
A 1.8 mm by 2.6 mm plate capacitor has the plates separated by a distance of 0.12 mm. (a) When a charge 4.00 × 10 −11 C of charge is placed on the capacitor, what is the electric field between the plates? (b) If the electric field with the dielectric constant 130 is placed between the plates while the charges on the capacitor stays the same, what is the dielectric field in the dielectric
Needs Complete typed solution with 100 % accuracy.
A parallel plate capacitor has metal plates, each of area 1.45 m², separated by 1.80 mm. What charge (in µC) is stored in this capacitor if a voltage of 2.70 × 103 v is applied to it?
The electric field strength is 1.90×104 N/C inside a parallel-plate capacitor with a 0.600 mm spacing. An electron is released from rest at the negative plate. What is the electron's speed when it reaches the positive plate?
A parallel plate capacitor consists of two rectangular, horizontally oriented plates, each with an area of 4.00 cm2, and separated by a distance of 0.300 mm. If the magnitude of the charge on each capacitor is 5.00 nC, what is the magnitude of the electric field between both plates?
An electron (-q = -1.602 x 10-19C) moves along a straight line from point a to point b with a separation distance d = 0.65m. Considering the electric field along this line is uniform with magnitude of 1.75x10^7 V/m and directed from point a to point b. What is the force on the electron? A) 2.4x10^-12 N B) -2.4x10^-12 N C) 2.8035x10^-12 N D) -2.8035x10^-12 N