1. Physics 121 - Electricity and Magnetism Lecture 3 - Electric Field Y&F Chapter 21 Sec. 4 – 7
Recap & Definition of Electric Field
Electric Field Lines
Charges in External Electric Fields
Field due to a Point Charge
Field Lines for Superpositions of Charges
Field of an Electric Dipole
Electric Dipole in an External Field: Torque and Potential Energy
Method for Finding Field due to Charge Distributions
Infinite Line of Charge
Arc of Charge
Ring of Charge
Disc of Charge and Infinite Sheet
Motion of a charged paricle in an Electric Field - CRT example

2. Review: Electric charge
Basics:
Positive and negative flavors. Like charges repel, opposites attract
Charge is conserved and quantized. e = 1.6 x Coulombs
Ordinary matter seeks electrical neutrality – screening
In conductors, charges are free to move around
screening and induction
In insulators, charges are not free to move around
but materials polarize
Coulombs Law: forces at a distance enabled by a field q1 q2
r12
F12
F21
3rd law pair of forces
Force on
q1 due to q2
(magnitude)
Constant
k=8.99x109 Nm2/coul2
repulsion
shown
Superposition of Forces or Fields
Point Charges

3. Fields (see also Lecture 01)
Temperature - T(r)
Pressure - P(r)
Gravitational Potential energy – U(r)
Electrostatic potential – V(r)
Electrostatic potential energy – U(r)
Scalar Field
Examples:
Fluid flow field (velocity v(r))
Gravitational field/acceleration - g(r)
Electric field – E(r)
Magnetic field– B(r)
Vector fields are “gradients” of scalar fields
Vector Field
Examples:
Gravitational Field versus Electrostatic Field
force/unit charge
q0 is a positive “test charge”
force/unit mass
m0 is a “test mass”
“Test” masses or charges map the directions and magnitudes of fields
Fields “explain” forces at a distance – space altered by source

4. Field due to a charge distribution
Test charge q0:
small and positive
does not affect the charge
distribution that produces E.
A charge distribution creates a field:
Map E field by moving q0 around and
measuring the force F at each point
E(r) is a vector parallel to F(r)
E field exists whether or not the
test charge is present
E varies in direction and magnitude
E does not depend on test charge q0 x y z
ARBITRARY
CHARGE
DISTRIBUTION
(PRODUCES E)
TEST q0
Most often…
F = Force on test charge q0 at point r due to the charge distribution
E = External electric field at point r = Force/unit charge
SI Units for E: Newtons / Coulomb later: V/m
Charge distributions can be: points, lines, surfaces, volumes

5. Magnitudes of Electrostatic Fields
Magnitude: E=F/q0
Direction: same as the force that acts on the positive test charge
SI unit: N/C
Field Location
Value
Inside copper wires in household circuits
10-2 N/C
Near a charged comb
103 N/C
Inside a TV or computer picture tube (CRT)
105 N/C
Near the charged drum of a photocopier
Breakdown E-field across an air gap (arcing)
3×106 N/C
E-field at the electron’s orbit in a hydrogen atom
5×1011 N/C
E-field on the surface of a Uranium nucleus
3×1021 N/C

6. Electric Field
3-1 A positive test charge of +q feels a force at a point P, due to an external electric field E pointing to the right. The test charge q is then replaced with a negative test charge of -q.
What changes happen to the external electric field at P and the force on the test charge?
A. The field and force both reverse direction
B. The force reverses direction, the field is unaffected.
C. The force is unaffected, the field is reversed
D. Both the field and the force are unaffected
E. The changes cannot be determined from the
information given.

7. Electric Field due to a point charge Q
Coulombs Law
test charge q0 Q r E q0
Find the electric field E due to point
charge Q as a function over all of space
Spherical symmetry: Magnitude E = KQ/r2 is constant on any spherical shell
Visualize: E field lines are radially out for Q > 0, in for Q < 0
Flux through any closed (spherical) shell enclosing Q is the same:
F = EA = Q.4pr2/4pe0r2 = Q/e Radius cancels
A closed (Gaussian) surface intercepts all the field lines leaving Q

8. Example: for point charges at r1, r2…..
Use superposition to calculate net electric field at each point due to a group of individual charges +
Example: for point charges at r1, r2…..
Do the sum above
for every test point i

9. Visualization: Electrostatic field lines (“Lines of force”)
Field lines begin on positive charges (or infinity and end on negative charges (or infinity).
Field lines cannot cross each other.
Map directions of electric field lines by moving a positive test charge around.
The tangent to a field line at a point shows the field direction there.
Where lines are dense the field is strong. The density of lines crossing a unit area perpendicular to the lines measures the strength of the field.
Repulsion
Attraction
WEAK
STRONG
Lines near a point charge

10. FIELD NEAR A “LARGE” (L>>d) SHEET OF UNIFORM + CHARGE
No conductor - just an infinitely large charge sheet on an insulator
The field has uniform intensity & direction everywhere
except on the sheet surface.
E approximately constant in the “near field” region (d << L) + + L d q F

11. Shell Theorem Conclusions
Field lines for other spherically symmetric charge distributions: spherical shell or solid sphere of charge
R is shell radius
Outside: r > R
Same field as point charge at center
Inside spherical distribution at distance r < R from center:
E = 0 for hollow shell;
E = kQinside/r2 for solid sphere (multiple shells)
Shell Theorem Conclusions

12. Example: Find Enet at a point on the axis of a dipole
Use superposition
Symmetry  Enet parallel to z-axis
z = 0
- q +q O z E+ E- r- r+ d
Exact
Limitation: z > d/2 or z < - d/2
DIPOLE MOMENT
points from – to +
Fields cancel better as d  0 so E 0
E falls off as 1/z3 not 1/z2
E is negative when z is negative
Does “far field” E look like point charge?
For z >> d : point “O” is “far” from center of dipole
Exercise: These formulas don’t describe E at the point midway between the charges
Ans: Emid = -4p/2pe0d3

13. Electric Field .B
3-2: Put the magnitudes of the electric field values at points A, B, and C shown in the figure in decreasing order.
A) EC>EB>EA
B) EB>EC>EA
C) EA>EC>EB
D) EB>EA>EC
E) EA>EB>EC .C .A

14. A Dipole in a Uniform EXTERNAL Electric Field What’s the net force
A Dipole in a Uniform EXTERNAL Electric Field What’s the net force? What’s the torque? (See Sec 21.7)
ASSSUME RIGID DIPOLE
Dipole Moment Vector
|torque| = 0 at q = 0 or q = p
|torque| = pE at q = +/- p/2
RESTORING TORQUE: t(-q) = t(+q)
OSCILLATOR
Fnet = 0
Torque = Force x moment arm
= - 2 q E x (d/2) sin(q)
= - p E sin(q)
(CW, into paper as shown)
Potential Energy U = -W
U = for q = +/- p/2
U = - pE for q = minimum
U = + pE for q = p maximum

15. 3-3: In the sketch, a dipole is free to rotate in a uniform external electric field. Which orientation has the smallest potential energy? E A B C D
3-4: Which orientation has the largest potential energy?

16. This process is just superposition
Method for finding the electric field at point P - - given a known continuous charge distribution
This process is just superposition
1. Find an expression for dq, the “point charge” within a differentially “small” chunk of the distribution
2. Represent field contributions at P due to a typical “point
charge” dq located in the distribution.
Use symmetry where possible.
3. Add up (integrate) the contributions dE over the whole distribution, varying the displacement and direction as needed.
Use symmetry where possible.

17. Example: Find electric field on the axis of a charged rod
Rod has length L, uniform positive charge per unit length λ, total charge Q. l = Q/L.
Calculate electric field at point P on the axis of the rod a distance a from one end. Field points along x-axis.
Add up contributions to the field from all locations of dq along the rod (x e [a, L + a]).
Limiting case a >> L or L/a  0
Rod approximates a point charge
E  point charge field formula

18. dq = lds = lRdq Electric field at center of an ARC of chargeP
dEp L
Uniform linear charge density l = Q/L
dq = lds = lRdq
P on symmetry axis at center of arc
 Net E is along y axis  need dEy only
Angle q varies between –q0 and +q0
Integrate:
In the plane of the arc
For a semi-circle, q0 = p/2
For a full circle, q0 = p

19. Electric field due to a RING of charge at point P on the symmetry (z) axis
SEE Y&F Example
21.9
Uniform linear charge density along circumference: l = Q/2pR
dq = lds = charge on arc segment of length ds = Rdf
P on symmetry axis  xy components of E cancel
Net E field is along z only, normal to plane of ring df
Integrate on azimuthal angle f from 0 to 2p
integral = 2p
Ez  0 as z  0
(see result for circular arc)
Exercise: Where is Ez a maximum?
Set dEz/dz = 0
Ans: z = R/sqrt(2)
Limit: For P “far away” use R << z
Ring approximates a point charge
if point P is very far away!

20. SEE Y&F
Example
21.10
Electric field of a charged straight LINE segment Point P on symmetry axis, a distance y off the line
uniform linear charge density: l = Q/L
point “P” is at y on symmetry axis
by symmetry, total E is along y-axis
x-components of dE pairs cancel
solve for line segment, then let y << L x y P L
dq = ldx r q
+q0
-q0 j ˆ r )
cos( dq k dE
kdq E d 2 y P q - =
“1 + tan2(q)” cancels in numerator and denominator
Find dx in terms of q:
Integrate from –q0 to +q0
Falls off as 1/y
For y << L (wire looks infinite) q0  p/2
Along –y direction
Finite length wire

21. Electric field due to a DISK of charge for point P on z (symmetry) axis
dEz f R x z q P r
dA=rdfdr s
See Y&F Ex 21.11
Surface charge density on disc is uniform: s = Q/pR2
Disc is a set of concentric rings, each with radial extent dr.
P on symmetry axis  net E field points only along z
dq = charge on area dA with arc rdf by radial extent dr
Integrate df first on azimuthal angle f from 0 to 2p yielding a factor of 2p
then integrate on ring radius r from 0 to R
Note Anti-derivative

22. “near field” is constant – disk looks like an infinite sheet of charge
Electric field due to a DISK of charge, continued
Exact Solution:
“near field” is constant – disk looks like an infinite sheet of charge
Near-Field: z<< R: P is “close” to the disk. Disk mimics an infinite sheet.
Far-Field: R<< z: P is “far” from the disk. Disk mimics a point charge.
Point charge formula

23. Near field approximates infinite uniformly charged sheet
Non-conductor, fixed surface charge density s
Method: solve non-conducting disc of charge for point on z-axis then approximate z << R (“near field”) L d
Infinite sheet  d<<L  use “near field” limit of disk
 uniform field

24. Motion of a Charged Particle in a Uniform Electric Field
A charge distribution produces E field at location
of charge q
Acceleration a is parallel or anti-parallel to E.
Acceleration is F/m not F/q = E
Acceleration is the same everywhere if field is uniform.
Example: Early CRT tube with electron gun and electrostatic deflector
ELECTROSTATIC
DEFLECTOR
PLATES
electrons are negative so acceleration a and electric force F are in the direction opposite the electric field E.
heated cathode (- pole) “boils off” electrons from the metal (thermionic emission)
FACE OF
CRT TUBE
ACCELERATOR
(electron gun controls intensity)

25. Motion of a Charged Particle in a Uniform Electric Field
Kinematics: ballistic trajectory
Dy is the DEFLECTION of the electron as it crosses the field
Acceleration has only a constant y component. vx is constant, ax=0 x Dy L vx
Measure deflection, find Dt via kinematics. Evaluate vy & vx
vx yields time of flight Dt
electrons are negative so acceleration a and electric force F are in the direction opposite the electric field E.
Use to find e/m