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Short Version : 21. Gauss’s Law
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21.1. Electric Field Lines
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.
Vector gives E at point
Field line gives direction of E
Spacing gives magnitude of E
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Field Lines of Electric Dipole
Direction of net field tangent to field line
Field is strong where lines are dense.
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Field Lines
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21.2. Electric Flux & Field
8 lines out of surfaces 1, 2, & 3. But 22 = 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.
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Number of field lines out of a closed surface net charge enclosed.
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 2.44 = 0 lines out of surface 3.
Count these.1: 42: 83: 44: 0
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Electric Flux
E A
Electric flux through flat surface A :
A flat surface is represented by a vector
where A = area of surface and
ˆAA a
ˆ / / normal of surfacea
[ ] N m2 / C.
surfaced E A
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.
E,
A,
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21.3. Gauss’s Law
Gauss’s law: The electric flux through any closed surface is proportional to the net charges enclosed.
d E A enclosedq depends on units.
For point charge enclosed by a sphere centered on it:
22
4q
k rr
q SI units
4 k 0
1
0
1
4 k
= vacuum permittivity12 2 28.85 10 /C N m
Gauss’s law: 0
enclosedqd
E A
Field of point charge: 2 20
ˆ ˆ4
q qkr r
E r r
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Gauss & Coulomb
For a point charge:
E r 2
A r 2
indep of r.
Gauss’ & Colomb’s laws are both expression of the inverse square law.
Principle of superposition argument holds for all charge distributions
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.
Outer sphere has 4 times area.But E is 4 times weaker.
So is the same
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21.4. Using Gauss’s Law
Useful only for symmetric charge distributions.
Spherical symmetry: r r ( point of symmetry at origin )
ˆE rE r r
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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.
ˆE rE r r
24 r E
0
Qr R
34
3
Q
R r
30
20
4
4
Q rr R
RE
Qr R
r
True for arbitrary spherical (r).
3
30
Q rr R
R
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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.
24 r E
0
0 r R
Qr R
2
0
0
4
r R
E Qr R
r
Contributions from A & B cancel.
Reflection symmetry E is radial.
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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.
24 r E
0
0
1
12
q r R
q q r R
20
20
4
4
qr R
rE
qr R
r
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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 = qq = 0
But E 0 on or inside
surface
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Line Symmetry
Line symmetry:
r r
ˆE r E r r
r = perpendicular distance to the symm. axis.
Distribution is independent of r// it must extend to infinity along symm. axis.
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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.
ˆE r E r r
2 r L E 0
L
02
Er
c.f. Eg. 20.7
True outside arbitrary radial (r).
(radial field)
No flux thru ends
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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.
2 r l E
0E
0
0 2.0
1 5.72.0
r cm
Cl r cm
L
0
0 2.0
15.7 2.0
2
r cm
EC r cm
L r
at r = 1.0 cm
9 2 2 612 9 10 5.7 10
3.0 0.03E N m C C
m m
at r = 3.0 cm
1.1 /M N C
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Plane Symmetry
Plane symmetry:
r r
ˆE r E r r
r = perpendicular distance to the symm. plane.
Distribution is independent of r// it must extend to infinity in symm. plane.
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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.
ˆE r E r r
2 A E 0
A
02
E
E > 0 if it points away from sheet.
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21.5. Fields of Arbitrary Charge Distributions
Dipole : E r 3 Point charge : E r 2
Line charge : E r 1
Surface charge : E const
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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.
point-charge-like
infinite-plane-charge-like
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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.
(a) Close to disk :02
E
71.44 10 /N C
(b) Far from disk :
2
QE k
R
24
2
Qk
r
22Qkr
99 2 2
2
20 102 9 10 /
0.005
CN m C
m
99 2 2
2
20 109 10 /
1.0
CN m C
m
180 /N C
14 /MN C
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21.6. Gauss’s Law & Conductors
Electrostatic Equilibrium
Conductor = material with free charges
E.g., free electrons in metals.
External E Polarization
Internal E
Total E = 0 : Electrostatic equilibrium
( All charges stationary )
Microscopic view: replace above with averaged values.
Neutral conductor Uniform field
Induced polarization cancels field inside
Net field
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Charged Conductors
Excess charges in conductor tend to
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.
= 0 thru this surface
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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.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
qout
+2C
qin
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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 1016 .
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Field at a Conductor Surface
At static equilibrium,
E = 0 inside conductor,
E = E at surface of conductor.
Gauss’ law applied to pillbox surface:
0
AE A
0
E
The local character of E ~ is incidental.
E always dependent on ALL the charges present.
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Dilemma?
E outside charged sheet of charge density was found to be02
E
E just outside conductor of surface charge density is0
E
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 = /20. Fields inside the conductor cancel, while those outside reinforce.
0
E
Hence, outside the conductor
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Application: Shielding & Lightning Safety
Coaxial cable
Strictly speaking, Gauss law applies only to static E.
However, e in metal can respond so quickly thathigh frequency EM field ( radio, TV, MW ) can also be blocked (skin effect).
Car hit by lightning,driver inside unharmed.
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Plate Capacitor
0
E
inside
0E outside
Charge on inner surfaces only