Short Version : 21. Gauss’s Law

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

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 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.

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

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,

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

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

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

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

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.

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

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

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.

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

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

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.

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.

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

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

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

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

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

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

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 .

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.

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

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.

Plate Capacitor

0

E

inside

0E outside

Charge on inner surfaces only