Medan Listrik Statis I (Hk.Couloumb, Hk. Gauss, Potensial Listrik) Sukiswo

ELECTROSTATICS - INTRODUCTION

ELECTRICAL CHARGES SOURCE of Electrostatic E-Field is CHARGE Examples of various charge distributions: 1. Point charge - Q (units of Coulomb) - model individual particle (eg. Electron) or a well-localized group of charge particles 2. Volume Charge Density -  or  v (units of Coulomb/m 3 ) - large # of particles - ignore discrete nature to smooth out distribution Eg. Doped Region of Semiconductor, e-beam in a cathode ray tube ( Beam has finite radius )

ELECTRICAL CHARGES 3. Surface Charge Density -  or  s (units of Coulomb/m 2 ) Other examples of Charge Distribution…….. Eg. Very thin charge layer on conductor surface 4. Line Charge Density -  or  l (units of Coulomb/m) - not as physically realizable Eg. Model for a wire, electron beam from far

ELECTRICAL CHARGES Maxwell’s equation: More generally, or Derived from:

COULOMB’S LAW (force), between point charges R Q1Q1 Q2Q2 Unit vector in r-direction Force on Charge 2 by Charge 1

COULOMB’S LAW - E Field,of Q 1 is Unit vector pointing away from Q 1 Then, - we work with E-Field because Maxwell’s equations written in those terms

E-FIELDS, is a VECTOR FieldHow do we represent it? Represent using Arrows : Direction and Length Proportional to Magnitude or strength of E-Field Point charge - Field points in the direction that a +ve test charge would move

E-FIELDS Computation of E-fields from multiple charges: Example: DIPOLE - 2 separated opposite polarity point charges +Q -Q x y +Q vector Resulting vector Apply superposition of Fields Planes of symmetry: Horizontal axis: E x cancels, E y adds Vertical axis: only E y component

E-FIELDS - Some examples

E-FIELDS - Dipole

E-FIELDS How would the DIPOLE field lines change if the charges were the same polarity? APPLICATION of SUPERPOSITION Usually text has many examples of setting up this integral In the course we will do some discrete  cases

ELECTROSTATICS - GAUSS’ LAW

MAXWELL’S FIRST EQUATION Differential FormIntegral Form Enclosed Charge - ‘dv’ integral over volume enclosed by ‘ds’ integral constant For vacuum and air - think of D and E as being the same D vs E depends on materials

Do Problem 1 Use Gauss’ Law to find D and E in symmetric problems GAUSS’ LAW - strategy Get D or E out of integral Always look at symmetry of the problem - and take advantage of this

GAUSS’ LAW - use of symmetry Example: A sheet of charge - charges are infinite in extent on say x,y plane x y z Arbitrary Point P Surface of infinite extent of charge, is sum due to all charges, points in all other components cancel only a function of z (not x or y) Can write down:

GAUSS’ LAW Do Problem 2a Problem 2b,is constant. For example a planar sheet of charge, where z is constant Problem 2cTo use GAUSS’ LAW, we need to find a surface that encloses the volume GAUSSIAN SURFACE - takes advantage of symmetry - when  is only a f(r) - when  is only a f(z)

GAUSS’ LAW Use Gaussian surface to “pull” this out of integral Integral now becomes: Usually an easy integral for surfaces under consideration

GAUSS’ LAW Example of using GAUSS’ law to find -a < z < a Z > a ; z< -a Z = a Z = -a “a slab of charge” By symmetry: If    > 0, then Z=0 From symmetry

GAUSS’ LAW First get in region |z| < a and create a surface at arbitrary z Use Gaussian surface with top at z = z’ and the bottom at -z’ Note: Gaussian Surface is NOT a material boundary

GAUSS’ LAW Evaluate LHS: =0, since These two integrals are equal

GAUSS’ LAW Key Step: Take E out of the Integral Computation of enclosed charge

GAUSS’ LAW (drop the prime)

GAUSS’ LAW Back to rectangular, slab geometry example….. Need to find, for |z| > a Z = -a Z = a  Z = -z’ Z = z’

GAUSS’ LAW As before, Computation of enclosed charge Note that the z-integration is from -a to a ; there is NO CHARGE outside |z|>a

GAUSS’ LAW Once again, For the region outside |z|>a

GAUSS’ LAW Plot of E-field as a function of z for planar example -a a z Note: E-field is continuous

ELECTROSTATICS - POTENTIALS

MAXWELL’S SECOND EQUATION Lesson 2.2 looked at Maxwell’s 1st equation: Today, we will use Maxwell’s 2nd equation: Importance of this equation is that it allows the use of Voltage or Electric Potential

POTENTIAL ENERGY Work done by a force is given by: If vectors are parallel, particle gains energy - Kinetic Energy If, Conservative Force Example : GRAVITY going DOWN increases KE, decreases PE going UP increases PE, decreases KE

POTENTIAL ENERGY If dealing with a conservative force, can use concept of POTENTIAL ENERGY For gravity, the potential energy has the form mgz Define the following integral: Potential Energy Change

POTENTIAL ENERGY Since and We can define: Potential Energy = Also define: Voltage = Potential Energy/Charge Voltage always needs reference or use voltage difference

POTENTIAL ENERGY Example: Use case of point charge at origin and obtain potential everywhere from E-field Spherical Geometry Point charge at (0,0,0) r Integration Path infinity Reference: V=0 at infinity

POTENTIAL ENERGY The integral for computing the potential of the point charge is: 0

POTENTIAL ENERGY - problems Do Problem 1a Hint for 1a: Use r=b as the reference - Start here and move away or inside r<b region R=a R=b R=r

POTENTIAL ENERGY - problems For conservative fields:,which implies that:, for any surface From vector calculus:,for any field f Can write: Define:

POTENTIAL SURFACES Potential is a SCALAR quantity Graphs are done as Surface Plots or Contour Plots Example - Parallel Plate Capacitor 0 +V 0 -V 0 +V 0 /2 -V 0 /2 -V 0 +V 0 E-Field Potential Surfaces

E-field from Potential Surfaces From: Gradient points in the direction of largest change Therefore, E-field lines are perpendicular (normal) to constant V surfaces Do problem 2 (add E-lines to potential plot)

Numerical Simulation of Potential In previous lesson 2.2, problem 3 and today in problem 1, Given  or Q E-fieldV derive Look for techniques so that V derive, given  or Q

Numerical Simulation of Potential For the case of a point charge: Distance from charge, is field point where we are measuring/calculating V, is location of charge

Numerical Simulation of Potential For smooth charge distribution: Volume charge distribution Line charge distribution

Numerical Simulation of Potential Problem 3 Setup for Problem 3a and 3b Line charge: Location of measurement of V origin Integrate along charge means dl is dz Line charge distribution

Numerical Simulation of Potential Problem 3 contd... Numerical Approximation Break line charge into 4 segments Charge for each segment Segment length Distance to charge

Numerical Simulation of Potential Problem 3 contd... For Part e…. Get V(r = 0.1) and V(r = 0.11) So..use 2 points to get  V and  r Use: V is a SCALAR field and easier to work with In many cases, easiest way to get E-field is to first find V and then use,