Electric Forces and Fields Joseph F. Alward, PhD Department of Physics University of the Pacific

                                                                                                             

The Atom:  A positively- charged nucleus surrounded by negatively-charged electrons.

Rubber scrapes electrons from fur atoms.

  Electrostatics:  the study of the behavior of stationary charges

    Electrostatic Attraction and Repulsion          

Some electrons from negatively charged rod move onto neutral ball.   Ball and rod are now both negatively charged.

Opposites attract. Likes repel.

         
   The Electroscope                                                                                                               

Two different electroscopes; one  uncharged, the other positively charged.

Hair separates for the same reason as do the gold leaves:  electrostatic repulsion of like charges.

 Attracting Uncharged Nonmetallic Objects                     

Charged rod polarizes atoms in paper.

Polarized Atom Electrons move to other side of atom, leaving the left side positively charged.

  Neutral Objects are Attracted to Charged Objects                   

Charged comb attracts neutral bits of paper.

Charged comb attracts neutral water molecules.

  Attracting Uncharged Metallic Objects      

Electrons are free to   move  in metals.

Nuclei remain in place;  electrons move to bottom.

  Applications of Electrostatic Charging                                        

Fine mist of negatively charged gold particles adhere to positively charged protein on fingerprint. (From Eugene Hecht's Physics, 2nd Edition Brooks/Cole Publishing)

Negatively charged paint adheres to positively charged metal.

  Electrostatic Air Cleaner   

  

Why are electrostatic phenomena, like shocks received from car door handles, more common in winter than in summer?

  Coulomb's Law                                      

Charles Augustin de Coulomb (1736-1806)  F = k Q1Q2 / r2

Q = charge in "Coulombs" (C)     r = charge separation  k  = 9 x 109         (SI units) ----------------------------------------- One electron has a charge of -1.6 x 10-19 C

  Important Numbers  

Charge of the electron:  -1.6 x 10-19 C = -e

Charge of the proton:     1.6 x 10-19 C = +e

Mass of the electron:      9.11 x 10-31 kg

Mass of the proton:  2000 times electron

  Simple Force Calculation.                                            

What is the force between the charges?  Is it a force of repulsion, or attraction?

F = k Q1Q2/r2  --------------------------------------  k = 9 x 109 N-m2/C2  F =   (9 x 109) (5)(8)/22     =   9 x 1010 N This is an enormous force, because a Coulomb is a huge charge:   One Coulomb is the charge on 6.25 x 1018 electrons.

    Board problems:  Equilateral triangle, rectangle.

  Three Charges on a Line:  Part I                                                                            .

Where may any test charge q be placed   between the charges if it is to experience   zero electric force?

Force between any two charges:  F = kq1q2/r2 ----------------------------------------------------------------  Forces by the two charges must  be equal but opposite: Force by red charge      =  k(5)q / x2 Force by yellow charge =  k(8)q / (4-x)2   Forces are equal:    k(5)q / x2 =  k(8)q / (4-x)2 Solve for x:     x = 1.77 m

 Three Charges on a Line: Part II                                                

On the line in which region,  A, B, or C, may a point be  found at which the net force  on a positive test charge q  would be zero? -------------------------------------------  F = kq1q2 / r2

  Charge Distributions                                                                                        

Charge on Metals  Metal Ball   Excess charge on   the surface of a   metal of uniform  curvature spreads  out.

Charge on Insulators   Plastic Ball  Charge on insulating  materials doesn't  move easily.

Charge on Metal Points Excess charge on a metal accumulates at points. Lightning, lightning rods.

  Charges Accumulate on Points                         

    Charging by Contact      

Some electrons leave rod and spread over sphere.

  Charging by Induction                                      

Rod does not touch sphere.  It pushes electrons out of the back side of the sphere and down the wire to ground.  The ground wire is disconnnected to prevent the return of the electrons from ground, then the rod is removed.

  The Electric Field Due to a Point Charge                                            

F = kQq0/r2        Define:  E = F/q0 = kQ/r2

Farther from the charge q0 the electric   field arrows are shorter.

  Electric Fields                                          

Electric field due to a positive point charge. Arrows point in the direction along which a positive test charge would accelerate. -------------------------------------------------------------------------  F = kQq0/r2                      E = F/q0 = kQ/r2

Electric field due to a negative point charge. ------------------------------------ Arrows point toward negative charge. Field is spherically symmetric.

  E-Field of Spherical Charge Distributions                      

E = kQ/r2 = (9x109)(5)/22                           = 1.125 x 1010 N/C

Radius of the ball is r = 0.5 m. What is the electric field E 2 m from the center of  the ball? (Assume uniform distribution)

 Electric Field Calculation                            

E2 = (3.0)2 + (2.0)2      = 13.0 E = 3.61 N/C q = tan-1(2/3)    = 33.7 degrees

   Symmetry In Electric Field  Calculations                                                                    

E = F/q0

 Electric Field of Dipoles                                           

    Electric Fields Under the Sea                                                          

Elephant Gnathonemus detects nearby objects by their effects on the electric field.

Cells in shark detect weak electric fields caused by the operation of the muscles of its prey.  Fields as weak as 10-6 N/C are detectable.

 Examples of Electric Field Strengths  

Source	E  (N/C)	Source	E  (N/C)
House wires	0.01	Thunderstorm	10,000
Near stereo	100	Breakdown of air	3 x 106
Atmosphere	150	Cell membrane	107
Shower	800	Laser	1011
Sunlight	1000	Pulsar	1014

  Compare to the field detectable by sharks,  10^-6 N/C.
   

    A Parallel Plate Capacitor                                                                  

s =  q/A = charge density E =  s/e0    e0 = 8.85 x 10-12 N-m2/C2 e0 is called the "permittivity of vacuum"

Example:   A = 0.15 m2   q  = 6 x 10-6 C   s  =  q/A       =  6 x 10-6 C/               0.15 m2       = 40 x 10-6 C/m2   E =  s/e0    = 40 x 10-6/          8.85 x 10-12    = 4.52 x 106 N/C

 Electric Field Inside a Conductor        

Excess charge inside a metal moves to the surface.

If E weren't zero inside, the free electrons (not shown) would accelerate.

   At equilibrium, all excess charge on a metal resides on the
   surface of the metal.

  Electric Fields Near Charged Conductors              

At equilibrium (electrostatic case), charges are stationary.  If the electric field weren't perpendicular to the metal, there would exist tangential forces on the electrons there, and the electrons on the surface would accelerate; this condition cannot exists if the electrons are stationary.

 E-Field is Perpendicular to Conductors in Equilibrium        

  Recognizing Incorrect Electric Field Patterns                                

This field configuration can't exist because the bottom of the ball will be positively charged, so a field should exist between the plate and the bottom of the ball.

On the left and right sides in this view, the electric field E is tangent to the metal ball, so a tangential force on the electrons would exist, contradicting the fact of equilibrium.

   Uncharged Metal Plate in an Electric Field                 

Metal plate is polarized by the external electric field. Sheets of charges on plate set up electric field (not shown) which cancels the external electric field. If the electric field E weren't zero inside the metal, what would happen?

  Using Metal to Shield Electronic Components                                   

Sensitive components are shielded from electricfields by surrounding them with metal. The metal pulls up from ground the charges necessary to cancel the electric field.

 Electric Flux Through a Plane Surface                    

Electric Flux = F = EA cos q

  Electric Flux Through a Closed Surface                

Electric Flux = F = E DA cos q (Some texts use DS for the area) ---------------------------------------------------- If there is no net charge inside this closed surface, the net flux is zero:   every arrow that enters must exit. E-field vectors which enter a surface provide negative flux, while vectors which exit give positive flux.

   Electric Flux Through Closed Surfaces                     

Four different closed  three- dimensional surfaces are indicated by broken lines. Flux through S1 is positive. Flux through S4 is negative. What are the fluxes through S2 and S3?

  Gauss's Law                                                                      

Friedrich Gauss (1777-1855) Gauss's Law: S AE cosq = q/e0

q = net charge inside Gaussian surface  This is useful if q = 0 and E = constant.

   Gauss's Law Gives Field Due to a Point Charge            

Gauss's Law:  SAE cosq = q/e0                A = area of sphere       = 4pr2   E  is the same at all points   on the surface         q = 0   cosq = 1   (4pr2)E = q/e0            E = q/(4pe0r2)

   Gauss's Law Application                                              

This is a sheet of charge--not a metal plate.  Sheet is very large (edges are not shown); near center of sheet, the E vector is perpendicular to the sheet.

SAE cosq = q/e0 q = sA    where s = charge density A1E + A2 (0) + A3E =  sA/e0 2AE =  sA/e0 E = s/2e0

 Gauss's Law Applied to Parallel Plate Capacitor                

Large plates close together; ignore   fringing at edges.  Electric field inside   the metal is zero.  E is perpendicular to   the plates (far from the edges).  We assume a charge density s.

E is zero at the left end and E is  parallel to the side.  q = sA      EA = sA/e0  E = s/e0