Download 20. Electric Charge, Force, & Field

21. Gauss’s Law
1.
2.
3.
4.
5.
6.
Electric Field Lines
Electric Flux & Field
Gauss’s Law
Using Gauss’s Law
Fields of Arbitrary Charge Distributions
Gauss’s Law & Conductors
Huge sparks jump to the operator’s cage, but the operator is unharmed. Why?
Ans: Gauss’ law  E = 0 inside cage
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.
8 lines out of surfaces 1 & 2.
16 lines out of surface 3.
0 out of 4.
16 lines out of surfaces 1, 2,
& 3. But 0 out of 4.
 8 lines out of surface 1.
8 lines out of surface 2.
44 = 0 lines out of surface 3.
8 lines out of surfaces 1, 2, & 3.
But 0 out of 4.
Count these.
Number of field lines out of a closed surface  net charge enclosed.
1:
2:
3:
4:
4
8
4
0
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.
GOT IT? 21.1.
The figure shows a cube of side s in a uniform electric field E.
What is the flux through each of the cube faces A, B, and C with the cube oriented as in (a) ?
Repeat for the orientation in (b), with the cube rotated 45 °.
A  0
A  0
B  0
2
 B   C  E s 2 cos 45  E s
C  E s 2
1
2
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.
GOT IT? 21.2.
A spherical surface surrounds an isolated positive charge, as shown.
If a 2nd charge is placed outside the surface,
which of the following will be true of the total flux though the surface?
(a)it doesn’t change;
(b)it increases;
Opposite charges
(c)it decreases;
Same charges
(d)it increases or decreases depending on the sign of the second charge.
Repeat for the electric field on the surface at the point between the charges.
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

0

rR
rR
 Qr
 4  R3 r  R

0
E
Q

rR
 4  0 r 2
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.
GOT IT? 21.3.
A spherical shell carries charge Q uniformly distributed over its surface.
If the charge on the shell doubles, what happens to the electric field
Stays 0.
(a) inside and
Doubles.
(b) outside
the shell?
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
r  2.0 cm


  2 r l E   1  5.7 C 
   L  l r  2.0 cm
 0

E 0
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

2AE 

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
GOT IT? 21.4.
A uniformly charged sheet measures 1m on each side,
and you are told the total charge Q.
What expression would you use to get approximate values for the field magnitude
(a) 1 cm from the sheet (but not near an edge) and
(b) 1 km from the sheet?

2 0
(a)
E
(b)
Ek
Q
r2

Q
A
r  1000 m
A  1 m2
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.
qin
+2C
qout
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
GOT IT? 21.5.
A conductor carries a net charge +Q.
There is a hollow cavity inside the conductor that contains a point charge Q.
In electrostatic equilibrium, is the charge on the outer surface of the conductor
(a) 2Q
(b) Q
(c) 0
(d) Q
(e) 2Q
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