Transcript Short Version : 21. Gauss’s Law

Short Version :
21. Gauss’s Law
21.1. Electric Field Lines
Vector gives
E at point
Field line gives
direction of E
Spacing gives
magnitude of E
Electric field lines = Continuous lines whose tangent is everywhere // E.
They begin at + charges & end at  charges or .
Their density is  field strength or charge magnitude.
Field Lines of Electric Dipole
Direction of net field
tangent to field line
Field is strong where
lines are dense.
Field Lines
21.2. Electric Flux & Field
8 lines out of surfaces 1, 2, & 3. But 22 = 0 out of 4 (2 out, 2 in).
16 lines out of surfaces 1, 2, & 3. But 0 out of 4.
8 lines out of (8 into) surfaces 1, 2, & 3. But 0 out of 4.
8 lines out of surfaces 1 & 2.
16 lines out of surface 3.
0 out of 4.
8 lines out of surface 1.
8 lines out of surface 2.
44 = 0 lines out of surface 3.
Count
these.
1: 4
2: 8
3: 4
4: 0
Number of field lines out of a closed
surface  net charge enclosed.
Electric Flux
A flat surface is represented by a vector A  A aˆ
where A = area of surface and aˆ / / normal of surface
Electric flux through flat surface A :
  E A
[  ]  N m2 / C.
E, 

surface
A, 
E dA
Open surface: can get from 1 side
to the other w/o crossing surface.
Direction of A ambiguous.
Closed surface: can’t get from 1
side to the other w/o crossing
surface.
A defined to point outward.
21.3. Gauss’s Law
Gauss’s law: The electric flux through any closed surface
is proportional to the net charges enclosed.


E  d A   qenclosed
 depends on units.
For point charge enclosed by a sphere centered on it:
k
  4 k 

0 
q
2
4

r
 q


2
r
1
4 k
1
0
 8.85 1012 C 2 / N  m2

E dA 
= vacuum permittivity
Ek
Field of point charge:
Gauss’s law:
SI units
qenclosed
0
q
q
ˆ
r

rˆ
2
2
r
4  0 r
Gauss & Coulomb
Outer sphere has 4 times area.
But E is 4 times weaker.
So  is the same
For a point charge:
E  r 2
Ar2
  indep of r.
Principle of superposition  argument
holds for all charge distributions
Gauss’ & Colomb’s laws are both
expression of the inverse square law.
For a given set of field lines going out of / into a point charge,
inverse square law  density of field lines  E in 3-D.
21.4. Using Gauss’s Law
Useful only for symmetric charge distributions.
Spherical symmetry:

 r     r 
E r   E  r  rˆ
( point of symmetry at origin )
Example 21.1. Uniformily Charged Sphere
A charge Q is spreaded uniformily throughout a sphere of radius R.
Find the electric field at all points, first inside and then outside the sphere.
E r   E  r  rˆ

 r  
Q
4 3
R
3
3
 Qr 
   R3 
 0 
2


  4 r E
 Q


rR
rR
0

 Qr
 4  R 3

0
E
Q

 4  0 r 2
rR
rR
True for arbitrary spherical (r).
Example 21.2. Hollow Spherical Shell
A thin, hollow spherical shell of radius R contains a total charge of Q.
distributed uniformly over its surface.
Find the electric field both inside and outside the sphere.
Reflection symmetry  E is
radial.
 0
  4 r 2 E  
 Q

 0

0


E
Q
 4  r 2
0

rR
rR
rR
rR
Contributions from A & B cancel.
Example 21.3. Point Charge Within a Shell
A positive point charge +q is at the center of a spherical shell of radius R
carrying charge 2q, distributed uniformly over its surface.
Find the field strength both inside and outside the shell.
1

rR
   q 

0
  4 r 2 E  
 1   q  2q  r  R
  0

q

 4  r 2

0
E
q

 4  0 r 2
rR
rR
Tip: Symmetry Matters
Spherical charge distribution inside a spherical shell is zero
 E = 0 inside shell
E  0 if either shell or distribution is not spherical.
Q = qq = 0
But E  0 on or inside surface
Line Symmetry
Line symmetry:
 r     r 
r = perpendicular distance to the symm. axis.
Distribution is independent of r//  it must extend to infinity along symm. axis.

E r   E  r  rˆ
Example 21.4. Infinite Line of Charge
Use Gauss’ law to find the electric field of an infinite line charge carrying
charge density  in C/m.
E r   E  r  rˆ
(radial field)
No flux thru ends

  2 r L E 

E

2  0 r
L
0
c.f. Eg. 20.7
True outside arbitrary radial (r).
Example 21.5. A Hollow Pipe
A thin-walled pipe 3.0 m long & 2.0 cm in radius carries a net charge q = 5.7 C
distributed uniformly over its surface.
Fine the electric field both 1.0 cm & 3.0 cm from the pipe axis, far from either end.
0


  2 r l E   1  5.7  C 
   L  l
 0

E0
r  2.0 cm
r  2.0 cm
0
r  2.0 cm


E
1
 2  L r  5.7 C  r  2.0 cm
0


at r = 1.0 cm
E  2   9 109 N m2 C 2 
5.7 10 C 

 3.0 m  0.03 m
1
6
 1.1 M N / C
at r = 3.0 cm
Plane Symmetry
 r     r 
Plane symmetry:
r = perpendicular distance to the symm.
plane.
Distribution is independent of r//  it must extend to infinity in symm. plane.

E r   E  r  rˆ
Example 21.6. A Sheet of Charge
An infinite sheet of charge carries uniform surface charge density  in C/m2.
Find the resulting electric field.
E r   E  r  rˆ

2 AE 

E
 A
0

2 0
E > 0 if it points away from sheet.
21.5. Fields of Arbitrary Charge Distributions
Dipole :
Line charge :
E  r 3
Point charge :
E  r 2
E  r 1
Surface charge :
E  const
Conceptual Example 21.1. Charged Disk
Sketch some electric field lines for a uniformly charged disk,
starting at the disk and extending out to several disk diameters.
infinite-plane-charge-like
point-charge-like
Making the Connection
Suppose the disk is 1.0 cm in diameter& carries charge 20 nC spread uniformly
over its surface.
Find the electric field strength
(a) 1.0 mm from the disk surface and
(b) 1.0 m from the disk.

E
2 0
(a) Close to disk :
 4 k
 2   9  10 N  m / C
9
2
2

Q
2  r 2 
20  109 C
 0.005 m 
2
 2k
Q
r2
 1.44 107 N / C
(b) Far from disk :
Q
Ek 2
R
  9  10 N  m / C
9
2
2

20  109 C
1.0 m 
2
 180 N / C
 14 MN / C
21.6. Gauss’s Law & Conductors
Electrostatic Equilibrium
Conductor = material with free charges
Neutral conductor
Uniform field
E.g., free electrons in metals.
Induced polarization
cancels field inside
External E  Polarization
 Internal E
Total E = 0 : Electrostatic equilibrium
( All charges stationary )
Microscopic view: replace
above with averaged values.
Net field
Charged Conductors
Excess charges in conductor tend to
 = 0 thru this surface
keep away from each other
 they stay at the surface.
More rigorously:
Gauss’ law with E = 0 inside conductor
 qenclosed = 0
 For a conductor in electrostatic equilibrium, all charges are on the surface.
Example 21.7. A Hollow Conductor
An irregularly shaped conductor has a hollow cavity.
The conductor itself carries a net charge of 1 C,
and there’s a 2 C point charge inside the cavity.
Find the net charge on the cavity wall & on the outer
surface of the conductor, assuming electrostatic
equilibrium.
qout
qin
+2C
E = 0 inside conductor
  = 0 through dotted surface
 qenclosed = 0
 Net charge on the cavity wall qin = 2 C
Net charge in conductor = 1 C = qout + qin
 charge on outer surface of the conductor qout = +3 C
Experimental Tests of Gauss’ Law
Measuring charge on ball is equivalent
to testing the inverse square law.
The exponent 2 was found to be
accurate to 1016 .
Field at a Conductor Surface
At static equilibrium,
E=0
inside conductor,
E = E at surface of conductor.
Gauss’ law applied to pillbox surface:
E A
 A
0

E

0
The local character of E ~  is incidental.
E always dependent on ALL the charges present.
Dilemma?
E outside charged sheet of charge density  was found to be E 
E just outside conductor of surface charge density  is
E

0
What
gives?
Resolution:
There’re 2 surfaces on the conductor plate.
The surface charge density on either surface is .
Each surface is a charge sheet giving E = /20.
Fields inside the conductor cancel, while those outside reinforce.
Hence,
E

0
outside the conductor

2 0
Application: Shielding & Lightning Safety
Coaxial cable
Car hit by lightning,
driver inside unharmed.
Strictly speaking, Gauss law applies only to static E.
However, e in metal can respond so quickly that
high frequency EM field ( radio, TV, MW ) can also be blocked (skin effect).
Plate Capacitor
E

0
E0
Charge on inner surfaces only
inside
outside