Bài giảng Vật lí Lớp 11 - Bài 19: Từ trường (Tiếng Anh)

nt r 
 = Force/unit charge 
SI Units for field: Newtons / Coulomb 
Electrostatic Field Examples 
Magnitude: E=F/q 0 
Direction: is that of 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 
10 3 N/C 
Inside a TV picture tube 
10 5 N/C 
Near the charged drum of a photocopier 
10 5 N/C 
Electric breakdown across an air gap 
3 × 10 6 N/C 
At the electron’s orbit in a hydrogen atom 
5 × 10 11 N/C 
On the surface of a Uranium nucleus 
3 × 10 21 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 × 10 6 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 
x 
y 
Find the force on a 1 m C . test charge at a point where the electrostatic field given by: 
The force depends on the field at the location of q 0 and nowhere else - 
E is due to a charge distribution somewhere; the distribution does not include q 0 . E may be constant or not. It is a vector. 
Solution: 
 F is in the same direction as E 
 If E is different at some other point 
 then F varies accordingly 
Electric Field due to a point charge Q 
Q 
r b 
r A 
E A 
E A 
E B 
E B 
E A 
E B 
Coulombs Law 
test charge q 0 
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/r 2 is constant on each spherical shell A & B in sketch 
 Flux through any closed shell enclosing Q is the same: 
 F A = E A A A = Q.4 p r 2 /4 pe 0 r 2 = Q/ e 0 
 and F B = E B A B = Q.4 p r 2 /4 pe 0 r 2 = Q/ e 0 
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 
TWO EQUAL + CHARGES (REPEL) 
EQUAL + AND – CHARGES ATTRACT 
WEAK 
STRONG 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
L 
d 
q 
F 
Field lines for a spherical shell  or solid sphere of charge 
Outside: 
 Same field as point charge 
Inside at r from center: 
 E = 0 for solid shell; E = kQ inside /r 2 for solid sphere 
Shell Theorem 
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 
DIPOLE MOMENT 
points from – to + 
Exercise: Find E at the point midway between the charges 
Ans : E = -4p/2 pe 0 d 3 
z = 0 
- q 
+q 
O 
z 
E + 
E - 
r - 
r + 
d 
Exact 
 falls off as 1/z 3 not 1/z 2 
 fields cancel as d  0 so E 0 
 negative when z is negative 
 “far field” region 
Assume d << z : point “O” is “far” from center of dipole 
Field lines for pairs of charges 
TWO EQUAL + CHARGES 
(REPEL – NOT A DIPOLE) 
EQUAL + AND – CHARGES 
(ATTRACT – A DIPOLE) 
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) E C >E B >E A 
 B) E B >E C >E A 
 C) E A >E C >E B 
 D) E B >E A >E C 
 E) E A >E B >E C 
. 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) 
 |torque| = 0 at q = 0 or q = p 
 |torque| = pE at q = +/- p /2 
 RESTORING TORQUE: t(-q ) = t(+q ) 
OSCILLATOR 
Potential Energy U = -W 
 U = - pE for q = 0 minimum 
 U = 0 for q = +/- p /2 
 U = + pE for q = p maximum 
3-3: In the sketch, a dipole is free to rotate in a uniform external electric field. Which configuration has the smallest potential energy? 
E 
A 
B 
C 
D 
E 
3-4: Which configuration has the largest potential energy? 
Method for finding the electric field at point Pgiven a known continuous charge distribution 
1. Find an expression for dq , 
 the charge in a “small” chunk of the distribution 
3. Add up (integrate) the contributions d E over the whole distribution, varying the displacement as needed 
2. Represent field contributions at P due to point 
 charges dq located in the distribution. Use symmetry 
Example: electric field along the axis of a charged rod 
Rod has length l with uniform positive charge per unit length λ and a total charge Q. l = Q/l. 
Calculate the electric field at a point P on the long axis of the rod a distance a from one end. Field points along x-axis. 
Interpret Limiting cases: 
l => 0 ? 
a >> l ? 
l >> a ? 
Add up contributions to the field from all locations of dq along the rod (x e [a, l + a]). 
Electric field due to LONG straight LINE of charge:P oint P on symmetry axis, a distance y off the line 
x 
y 
P 
L 
dq = l dx 
r 
q 
+q 0 
-q 0 
 “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 + tan 2 ( q )” cancels in numerator and denominator 
 Integrate from – q 0 to + q 0 
Falls off as 1/y 
 As q 0  p /2 LONG wire looks infinite 
Along –y direction 
Falls off as 1/y 2 
for y>>L 
Find dx in terms of q 
Infinite ( i.e.”large ”) uniformly charged sheet 
Non-conductor, fixed surface charge density s 
L 
d 
infinite sheet  d<<L  “near field”  uniform field 
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 = l dl = l Rd q 
+q 0 
-q 0 
q 
R 
dq 
P 
d E 
L 
 P on symmetry axis at center of arc 
  net E is along y axis: E y only 
 Angle q is between – q 0 and + q 0 
 Integrate: 
 For a semi-circle, q 0 = p /2 
 For a full circle, q 0 = p 
Electric field due to a RING of charge for point P on the symmetry axis 
 Uniform linear charge density along circumference 
 l = Q/2 p R 
 P on symmetry axis  net E field only along z 
 dq = charge on arc segment ds 
d f 
 Integrate on azimuthal angle f from 0 to 2 p 
integral = 2 p 
E  0 as z  0 
(see result for arc) 
Exercise: Where is E z a maximum? 
Set dE z /dz = 0 
 Ans : z = R/sqrt(2) 
 Limit: For P “far away” use z >> R 
Ring looks like a point charge 
if point P is very far away! 
Electric field due to a DISK of chargefor point P on z (symmetry) axis 
d E 
f 
R 
x 
z 
q 
P 
r 
dA = rdrd f 
s 
 Uniform surface charge density on disc in x-y plane 
 s = Q/ p R 2 
 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 
 Integrate twice: first on azimuthal angle f from 0 to 2 p which yields a factor of 2p 
 then on ring radius r from 0 to R 
Note Anti-derivative 
Field is constant near an infinite sheet of charge 
For z<< R : disc looks infinitely large if P is close to the surface 
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 
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 “boils off” electrons from the metal ( thermionic emission) 
 FACE OF 
CRT TUBE 
 ELECTROSTATIC 
ACCELERATOR 
PLATES 
(electron gun controls intensity) 
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. 
x 
D y 
L 
v x 
FIND DEFLECTION D y of the electron as it crosses the field 
 Acceleration has only y component 
 v x is constant 
Find deflection and vertical speed via kinematics 
 Time of flight 
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