Lecture 11: Potential Gradient and Capacitor Review:

Similar documents
Electric Potential. and Equipotentials

q r 1 4πε Review: Two ways to find V at any point in space: Integrate E dl: Sum or Integrate over charges: q 1 r 1 q 2 r 2 r 3 q 3

Electric Field F E. q Q R Q. ˆ 4 r r - - Electric field intensity depends on the medium! origin

Electricity & Magnetism Lecture 6: Electric Potential

U>, and is negative. Electric Potential Energy

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface

(A) 6.32 (B) 9.49 (C) (D) (E) 18.97

Physics 604 Problem Set 1 Due Sept 16, 2010

Physics 11b Lecture #11

PX3008 Problem Sheet 1

ELECTROSTATICS. 4πε0. E dr. The electric field is along the direction where the potential decreases at the maximum rate. 5. Electric Potential Energy:

FI 2201 Electromagnetism

(1) It increases the break down potential of the surrounding medium so that more potential can be applied and hence more charge can be stored.

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

Continuous Charge Distributions

Chapter 25 Electric Potential

ELECTRO - MAGNETIC INDUCTION

Physics 1402: Lecture 7 Today s Agenda

Physics 1502: Lecture 2 Today s Agenda

Ch 26 - Capacitance! What s Next! Review! Lab this week!

General Physics (PHY 2140)

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

Physics 505 Fall 2005 Midterm Solutions. This midterm is a two hour open book, open notes exam. Do all three problems.

Chapter 28 Sources of Magnetic Field

Chapter 23 Electrical Potential

r = (0.250 m) + (0.250 m) r = m = = ( N m / C )

This immediately suggests an inverse-square law for a "piece" of current along the line.

CHAPTER 25 ELECTRIC POTENTIAL

Solutions to Midterm Physics 201

CHAPTER 18: ELECTRIC CHARGE AND ELECTRIC FIELD

π,π is the angle FROM a! TO b

Answers to test yourself questions

Review for Midterm-1

= ΔW a b. U 1 r m 1 + K 2

Homework Assignment 3 Solution Set

Chapter 22 The Electric Field II: Continuous Charge Distributions

Prof. Anchordoqui Problems set # 12 Physics 169 May 12, 2015

Work, Potential Energy, Conservation of Energy. the electric forces are conservative: ur r

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3

CAPACITORS AND DIELECTRICS

Chapter 4. Energy and Potential

Energy Dissipation Gravitational Potential Energy Power

The Formulas of Vector Calculus John Cullinan

This chapter is about energy associated with electrical interactions. Every

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

Chapter 2: Electric Field

PHYS 2421 Fields and Waves

Physics 2135 Exam 1 February 14, 2017

Course Updates. Reminders: 1) Assignment #8 available. 2) Chapter 28 this week.

ELECTROSTATICS. Syllabus : Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road PE 1

2 mv2 qv (0) = 0 v = 2qV (0)/m. Express q. . Substitute for V (0) and simplify to obtain: v = q

Today s Plan. Electric Dipoles. More on Gauss Law. Comment on PDF copies of Lectures. Final iclicker roll-call

Designing Information Devices and Systems I Discussion 8B

SOLUTIONS TO CONCEPTS CHAPTER 11

PHYS 1444 Lecture #5

( ) D x ( s) if r s (3) ( ) (6) ( r) = d dr D x

MAGNETIC EFFECT OF CURRENT & MAGNETISM

Fluids & Bernoulli s Equation. Group Problems 9

Gauss Law. Physics 231 Lecture 2-1

7.2.3 Inductance. Neumann Formula for the Mutual Inductance. Important Things about Mutual Inductance

Chapter 21: Electric Charge and Electric Field

dx was area under f ( x ) if ( ) 0

10 Statistical Distributions Solutions

SPA7010U/SPA7010P: THE GALAXY. Solutions for Coursework 1. Questions distributed on: 25 January 2018.

Chapter 4 Kinematics in Two Dimensions

Chapter 24. Gauss s Law

Chapter 25: Current, Resistance and Electromotive Force. ~10-4 m/s Typical speeds ~ 10 6 m/s

Module 05: Gauss s s Law a

Chapter 4 Two-Dimensional Motion

Reading from Young & Freedman: For this topic, read the introduction to chapter 24 and sections 24.1 to 24.5.

10 m, so the distance from the Sun to the Moon during a solar eclipse is. The mass of the Sun, Earth, and Moon are = =

2.2 This is the Nearest One Head (Optional) Experimental Verification of Gauss s Law and Coulomb s Law

Class Summary. be functions and f( D) , we define the composition of f with g, denoted g f by

of Technology: MIT OpenCourseWare). (accessed MM DD, YYYY). License: Creative Commons Attribution- Noncommercial-Share Alike.

MATHEMATICS IV 2 MARKS. 5 2 = e 3, 4

Lecture 4. Electric Potential

Physics Jonathan Dowling. Lecture 9 FIRST MIDTERM REVIEW

( ) ( ) ( ) ( ) ( ) # B x ( ˆ i ) ( ) # B y ( ˆ j ) ( ) # B y ("ˆ ( ) ( ) ( (( ) # ("ˆ ( ) ( ) ( ) # B ˆ z ( k )

Today in Physics 122: work, energy and potential in electrostatics


Topics for Review for Final Exam in Calculus 16A

3.1 Magnetic Fields. Oersted and Ampere

Unit 1. Electrostatics of point charges

1. Viscosities: μ = ρν. 2. Newton s viscosity law: 3. Infinitesimal surface force df. 4. Moment about the point o, dm

Hopefully Helpful Hints for Gauss s Law

Chapter 25: Current, Resistance and Electromotive Force. Charge carrier motion in a conductor in two parts

Problem Set 3 SOLUTIONS

$ i. !((( dv vol. Physics 8.02 Quiz One Equations Fall q 1 q 2 r 2 C = 2 C! V 2 = Q 2 2C F = 4!" or. r ˆ = points from source q to observer

CHAPTER 2 ELECTROSTATIC POTENTIAL

Continuous Charge Distributions: Electric Field and Electric Flux

Algebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016

Friedmannien equations

Resistors. Consider a uniform cylinder of material with mediocre to poor to pathetic conductivity ( )

Welcome to Physics 272

6. Gravitation. 6.1 Newton's law of Gravitation

Polynomials and Division Theory

Collection of Formulas

Homework: Study 6.2 #1, 3, 5, 7, 11, 15, 55, 57

Example 2: ( ) 2. $ s ' 9.11" 10 *31 kg ( )( 1" 10 *10 m) ( e)

Question Bank. Section A. is skew-hermitian matrix. is diagonalizable. (, ) , Evaluate (, ) 12 about = 1 and = Find, if

Transcription:

Lectue 11: Potentil Gdient nd Cpcito Review: Two wys to find t ny point in spce: Sum o Integte ove chges: q 1 1 q 2 2 3 P i 1 q i i dq q 3 P 1 dq xmple of integting ove distiution: line of chge ing of chge disk of chge

Detemine fom : Detemining fom : Δ d lecticl Potentil Fo n infinitesiml step: dl dl xmple: due to spheicl chge distiution. dl dl dl cosθ Cses: θ : d - dl θ 9 o : d θ 18 o : d - dl(-1) dl Cn wite: d dl ( ( x dx + y dy iˆ + x + z y ˆj + dz) z θ dl d kˆ) ( dxiˆ + dy diectionl deivtive d depends on diection mximum chnge fo θ o 18 degees F θ dl ˆj + dzkˆ)

Potentil Gdient Tke step in x diection: (dy dz ) Similly: And: d ( dx + dy + dz) x y x d dx y y, z const. y z x z ( iˆ + ˆj + x y z ˆ ( ˆ i + j + kˆ) x y z z kˆ) x dx gdient opeto Gdient of points in the diection tht inceses the fstest with espect to chnge in x, y, nd z. points in the diection tht deceses the fstest. pependicul to equilpotentil lines. ptil deivtive

Potentil Gdient xmple: chge in unifom field U qy U/q y whee is tken s t y. ( iˆ + x (iˆ + ˆj + kˆ) ˆ j + y z ˆj kˆ) y Given o in some egion of spce, cn find the othe. Cylindicl nd spheicl symmety cses: Fo dil cse nd is distnce fom point (spheicl) o xis (cylindidl): o y q xmple: of point chge: q ( ) q 1 q ( )( ) 2 2

xmple: The electic potentil in egion of spce is given y (x, y, z) A(x 2 3y 2 + z 2 ) whee A is constnt. Deive n expession fo the electic field t ny point in this egion. ( x iˆ + y ˆj + z kˆ){ A( x 2 3y 2 + z 2) )} (2Axiˆ 6Ay ˆj + 2Azkˆ) 2A( xiˆ 3yj ˆ + zkˆ)

UI7PF7: This gph shows the electic potentil t vious points long the x-xis. 2) At which point(s) is the electic field zeo? A B C D

UI7ACT1 1 The electic potentil in egion of spce is given y ( ) 2 3 x 3x x The x-component of the electic field x t x 2 is () x () x > (c) x <

1 UI7ACT1 The electic potentil in egion of spce is given y 2 3 ( x) 3x x The x-component of the electic field x t x 2 is () x () x > (c) x < We know (x) eveywhee To otin x eveywhee, use x x x x 6x + 3x ( 2 ) 12 + 12 2

CAPACITOR A cpcito is device fomed with two o moe septed conductos tht stoe chge nd electic enegy. Conside ny two conductos nd we put + on nd on. Conducto hs constnt nd conducto hs constnt, then dl Y&F fig. 24.1 The electic field is popotionl to the chges ±. If we doule the chges ±, the electic field doules. Then the voltge diffeence is - popotionl to the chge. This popotionlity depends on size, shpe nd seption of the conductos. const ( )

CAPACITOR, continued If we cll this constnt, Cpcitnce, C, nd the voltge diffeence, -, then, C O C Cpcitnce, depends on the geomety of the two conductos (size, shpe, seption) nd cpcitnce is lwys positive quntity y its definition (voltge diffeence nd chge of + conducto) UNITs of cpcitnce, Coulom/olts o Fds, fte Michel Fdy

xmple: Pllel Plte Cpcito (idel) Clculte the cpcitnce. We ssume +σ, -σ chge densities on ech plte with potentil diffeence : C d A + + + + ----- Need : σa Need : fom def n: Use Guss Lw to find dl

Recll: Two Infinite Sheets (into sceen) Field outside the sheets is zeo Gussin sufce encloses zeo net chge Field inside sheets is not zeo: Gussin sufce encloses non-zeo net chge σa ds A inside σ ε σ σ A A -

xmple: Pllel Plte Cpcito Clculte the cpcitnce: Assume +, - on pltes with potentil diffeence. σ ε Aε d A + + + + dl ----- dl d A ε d Aε d C As expected, the cpcitnce of this cpcito depends only on its geomety (A,d). Note tht C ~ length; this will lwys e the cse!

Cylindicl Cpcito xmple Clculte the cpcitnce: Assume +, - on sufce of cylindes with potentil diffeence. Gussin sufce is cylinde of dius ( < < ) nd length L - + L Apply Guss' Lw: ds 2πL ε 2πε L If we ssume tht inne cylinde hs +, then the potentil is positive if we tke the zeo of potentil to e defined t : dl d d πε L 2πε L 2 ln C 2 πε ln L

Spheicl Cpcito xmple Suppose we hve 2 concentic spheicl shells of dii nd nd chges + nd. uestion: Wht is the cpcitnce? etween shells is sme s point chge +. (Guss s Lw): + - 1 2 o dl dl C d 1 1 ( o ) d 2 d C 1 ( 1 )

Summy A Cpcito is n oject with two sptilly septed conducting sufces. The definition of the cpcitnce of such n oject is: C The cpcitnce depends on the geomety : d A + + + + ----- + - L - + Pllel Pltes Aε d C C Cylindicl 2 πε L ln C Spheicl