Bài giảng về Điện trường (tiếng anh)

Electricity and Magnetism
Lecture 3 - Electric Field
Physics 121 - Chapter 23 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 (Ch. 26.6)
Method for Finding Field due to Charge Distributions
Infinite Line of Charge
Arc of Charge
Ring of Charge
Disc of Charge and Infinite Sheet
Crossed Electric Fields: CRT example
Summary
Fields (see lecture 01)
Scalar Field
Examples:
Temperature - T(r)
Pressure - P(r)
Potential energy – U(r)
Electrostatic potential – V(r)
Vector Field
Examples:
Velocity - v(r)
Gravitational field/acceleration - g(r)
Electric field – E(r)
Magnetic field– B(r)
Fields “explain” forces at a distance – space altered by source
Definition: Electric field and the force due to it
Test charge concept:
assume small and positive
does not affect the charge distribution
The charge distribution creates the field
The E field exists whether or not the
test charge is present
E(r) is a vector parallel to F(r)
Map the E field by moving q0 around. E varies
in direction and magnitude
most often…
Electrostatic Field Examples
Magnitude: E=F/q0
Direction: is that of the force that acts on the positive test charge
SI unit: N/C
Electric Field
3-1 A test charge of +3 µC is at a point P where there is an external electric field pointing to the right whose magnitude is 4×106 N/C. The test charge is then replaced with another test charge of –3 µC.

What happens to the external electric field at P and the force on the test charge when the change happens?

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
Example: Calculating force on a charge in an electric field
Find the force on a 1 mC. test charge at a point where the electrostatic field given by:
Electric Field due to a point charge Q
Coulombs Law
test charge q0
Find the field E due to point
charge Q as a function over
all of space
The direction of E is radial: out for +|Q|, in for -|Q|
Magnitude of E = KQ/r2 is constant on each spherical shell A & B in sketch
Flux through any closed shell enclosing Q is the same:
FA = EAAA = Q.4pr2/4pe0r2 = Q/e0
and FB = EBAB = Q.4pr2/4pe0r2 = Q/e0
The closed surfaces intercept all the field lines
Visualization: electric field lines (lines of force)
The direction of an electric field line is mapped out by moving a positive test charge around in the field.
The tangent to a field line shows its direction at that point.
The density of lines crossing a unit area perpendicular to the lines measures the strength of the field. Lines are close together where the electric field is strong and far apart where it is weak.
Lines begin on positive charges (or infinity and end on negative charges (or infinity).
Lines cannot cross
DETAIL NEAR A POINT CHARGE
no conductor - just an infinitely large charge sheet
good approximation in the “near field” region (d << L)
The field has
uniform intensity
& direction everywhere
NEAR A LARGE, UNIFORM SHEET OF + CHARGE
Field lines for a spherical shell
or solid sphere of charge
Use superposition to calculate net electric field at each point due to a group of individual charges
Do this sum for every test point
Example: point charges
Example: electric field on the axis of a dipole
Find E at point “O” due to DIPOLE
Use superposition, assume z >> d
Symmetry  E parallel to z-axis
Exercise: Find E at the point midway between the charges
Ans: E = -4p/2pe0d3
Field lines for pairs of charges
Field lines map out the SUPERPOSED fields
Electric Field
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
.B
Electric dipole in uniform EXTERNAL electric field (Ch. 26.6)
- feels torque - stores potential energy
ASSSUME THE DIPOLE IS A RIGID OBJECT
Torque = Force x moment arm
= - 2 q E x (d/2) sin(q)
= - p E sin(q)
(CW, into paper as shown)
3-3: In the sketch, a dipole is free to rotate in a uniform external electric field. Which configuration has the smallest potential energy?
3-4: Which configuration has the largest potential energy?
Method for finding the electric field at point P
given a known continuous charge distribution
1. Find an expression for dq,
the charge in a “small” chunk of the distribution
Example: electric field along the axis of a charged rod
Electric field due to LONG straight LINE of charge:
Point P on symmetry axis, a distance y off the line
“LONG” means y << L
point “P” is at y on symmetry axis
linear charge density: l = Q/L
by symmetry, total E is along y-axis
x-components of dE pairs cancel
“1 + tan2(q)” cancels in numerator and denominator
Infinite (i.e.”large”) uniformly charged sheet
Non-conductor, fixed surface charge density s
L
d
infinite sheet  d<method: solve non-conducting disc of charge for point on z-axis then approximate z << R
Electric field due to an ARC of charge, at center of arc
Uniform linear charge density l = Q/L
dq = ldl = lRdq
P on symmetry axis at center of arc
 net E is along y axis: Ey only
Angle q is between –q0 and +q0
Electric field due to a RING of charge
for point P on the symmetry axis
Uniform linear charge density along circumference
l = Q/2pR
P on symmetry axis  net E field only along z
dq = charge on arc segment ds
df
Electric field due to a DISK of charge
for point P on z (symmetry) axis
Uniform surface charge density on disc in x-y plane
s = Q/pR2
Disc is a set of rings, each of them dr wide in radius
P on symmetry axis  net E field only along z
dq = charge on arc segment ds and radial segment dr
Motion of a Charged Particle in a Uniform Electric Field
Stationary charges produce 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
Motion of a Charged Particle in a Uniform Electric Field
electrons are negative so acceleration a and electric force F are in the direction opposite the electric field E.
FIND DEFLECTION Dy of the electron as it crosses the field
Acceleration has only y component
vx is constant

Stationary charges produce 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
Lecture 3
Summary: Chapter 22: Electric Field
Electric field on equator of a DIPOLE using vector notation
EXTRA
SUMMARY: RESULTS of electric field calculations
EXTRA
Charge distributions comparison
linear, surface, volume
EXTRA
EXTRA
EXTRA