PHYSICS 182 Lecture Notes - Lecture 20: Electric Potential, Electric Field, Voltage

33 views3 pages

cerisebuffalo390

4 May 2017

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Capacitors with dielectrics in combination: capacitors a and b have identical plate areas and gap separations. The space between the plates of each capacitor is half-filled with a dielectric of dielectric constant k as shown. Find the capacitance of each in terms of k and c0 (original capacitor) Capacitors in combination: two identical parallel plate capacitors are connected to a battery in (a) parallel and (b) series. Electric current: current measure of flow of charge (+ charge, by convention, counts total charge dq passing through area in a time interval dt. In electronic circuits, electrons ( e) carry current, flow opposite to current. In liquid or gas, both cations and anions can carry current. Electric current: current flows in conductors, where the wire is long, thin piece of conductor (metal) It is no longer is equilibrium, need not have e = 0.

Related Questions

A parallel-plate capacitor has plates of area A = 7.00 X 10^-2 m²? separated by distance d = 2.00 X 10^-4 m. (a) Calculate the capacitance if the space between the plates is filled with air. What is the capacitance if the space is filled half with air and half with a dielectric constant k = 3.70 as in (b) Figure a, and (c) Figure b? (Hint: In (b) and (c), one of the capacitors is a parallel combination and the other is a series combination.)

raspberryhorse508

A parallel plate capacitor of capacitance 6.0 μF has the space between the plates filled with a slab of glass with k = 3.0. The capacitor is charged by attaching it to a 1.5-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed. Find (a) the capacitance, (b) the potential difference, (c) the charge on the plates, and (d) the energy stored in the capacitor after the glass is removed.

To understand the meaning of capacitance and ways of calculating capacitance.

When a positive charge q is placed on a conductor that is insulated from ground, an electric field emanates from the conductor to ground, and the conductor will have a nonzero potential V relative to ground. If more charge is placed on the conductor, this voltage will increase proportionately. The ratio of charge to voltage is called the capacitance C of this conductor: C=q/V.

Capacitance is one of the central concepts in electrostatics, and specially constructed devices called capacitors are essential elements of electronic circuits. In a capacitor, a second conducting surface is placed near the first (they are often called electrodes), and the relevant voltage is the voltage between these two electrodes.

This tutorial is designed to help you understand capacitance by assisting you in calculating the capacitance of a parallel-plate capacitor, which consists of two plates each of area A separated by a small distanced with air or vacuum in between. In figuring out the capacitance of this configuration of conductors, it is important to keep in mind that the voltage difference is the line integral of the electric field between the plates.

Assume that charge -q is placed on the top plate, and +q is placed on the bottom plate. What is the magnitude of the electric field E between the plates?
What is the voltage V between the plates of the capacitor?

beigemouse1

PHYSICS 182 Lecture Notes - Lecture 20: Electric Potential, Electric Field, Voltage

Document Summary

Get access

Related textbook solutions

Physics: Principles with Applications

Related Documents

PHYSICS 182 Lecture Notes - Lecture 16: Farad, Electric Field

PHYSICS 5C Chapter 23: electrical potential

PHYS 142 Chapter Notes - Chapter 26: Wax Paper, Electrolytic Capacitor, Dielectric

Related Questions