Physics 2113 Lecture: 09 Gauss’ Law II

Flux Capacitor (Schematic)

Physics 2113 Jonathan Dowling

Carl Friedrich Gauss 1777 – 1855

Gauss’ Law: General Case

•  Consider any ARBITRARY CLOSED surface S -- NOTE: this does NOT have to be a “real” physical object!

•  The TOTAL ELECTRIC FLUX through S is proportional to the TOTAL CHARGE ENCLOSED!

•  The results of a complicated integral is a very simple formula: it avoids long calculations!

Φ ≡

!E ⋅d!A = qins

ε0Surface"∫

S

(One of Maxwell’s 4 equations!)

qins

Gauss’ Law: ICPP

"+ "= +q"− "= −q

Φ ≡!E ⋅d!A =

S2"∫

+q / ε0 ?−q / ε0 ?0?

Φ ≡!E ⋅d!A =

S3"∫

+q / ε0 ?−q / ε0 ?0?

Φ ≡!E ⋅d!A =

S4"∫

+q / ε0 ?−q / ε0 ?0?

Φ ≡!E ⋅d!A =

S1"∫

+q / ε0 ?−q / ε0 ?0?

Φ1 = +2 + 5 + 7 − 3− 4 − 7 Φ2 = +3+ 5 +10 − 3− 4 − 6 Φ3 = +2 + 5 + 8 − 5 − 6 − 7

+ −0

23-3 A Charged Isolated Conductor

•  23.14 Apply the relationship between surface charge density σ and the area over which the charge is uniformly spread.

•  23.15 Identify that if excess charge (positive or negative) is placed on an isolated conductor, that charge moves to the surface and none is in the interior.

•  23.16 Identify the value of the electric field inside an isolated conductor.

•  23.17 For a conductor with a cavity that contains a charged object, determine the charge on the cavity wall and on the external surface.

•  23.18 Explain how Gauss’ law is used to find the electric field magnitude E near an isolated conducting surface with a uniform surface charge density σ.

•  23.19 For a uniformly charged conducting surface, apply the relationship between the charge density σ and the electric field magnitude E at points near the conductor, and identify the direction of the field vectors.

Learning Objectives

Properties of Conductors Inside a Conductor in Electrostatic Equilibrium, the Electric Field Is ZERO. Why? Because If the Field Is Not Zero, Then Charges Inside the Conductor Would Be Moving. SO: Charges in a Conductor Redistribute Themselves Wherever They Are Needed to Make the Field Inside the Conductor ZERO. Excess Charges Are Always on the Surface of the Conductors.

ICPP: Conducting Sphere •  A spherical conducting shell has

an excess charge of +10 C. •  A point charge of –15 C is

located at center of the sphere. •  Use Gauss’ Law to calculate the

charge on inner and outer surface of spherical shell

(a) Inner: +15 C; outer: 0 (b) Inner: 0; outer: +10 C (c) Inner: +15 C; outer: –5 C

–15C

R2 R1

+10 C

"+ "= +5C"− "= −5C

++

+

−

−

−

+ +

Hint: E-Field is Zero inside conductor so

!E ⋅d!A =

S1"∫ 0 = −15C / ε0 + ?

S1

S2

Gauss’ Law: Conducting Sphere •  Inside a conductor, E = 0 under

static equilibrium! Otherwise electrons would keep moving!

•  Construct a Gaussian surface inside the metal as shown. (Does not have to be spherical!)

–15C •  Since E = 0 inside the metal, flux through this surface = 0

•  Gauss’ Law says total charge enclosed = 0

•  Charge on inner surface = +15 C

Since TOTAL charge on shell is +10 C, Charge on outer surface = +10 C - 15 C = -5 C!

–5 C

+15C

Faraday’s Cage •  Given a hollow conductor of arbitrary shape.

Suppose an excess charge Q is placed on this conductor. Suppose the conductor is placed in an external electric field. How does the charge distribute itself on outer and inner surfaces?

(a) Inner: Q/2; outer: Q/2 (b) Inner: 0; outer: Q (c) Inner: Q; outer: 0

Inside cavity is “shielded” from all external electric fields! “Faraday Cage effect”

• Choose any arbitrary surface inside the metal • Since E = 0, flux = 0 • Hence total charge enclosed = 0 All charge goes on outer surface!

Safe in the Plane!?

Faraday’s Cage: Electric Field Inside Hollow Conductor is Zero

Inside cavity is “shielded” from all external electric fields! “Faraday Cage effect”

• Choose any arbitrary surface inside the metal • Since E = 0, flux = 0 • Hence total charge enclosed = 0 All charge goes on outer surface!

Safe in the Plane!?

Safe in the Car!?

E=0

Field on Conductor Perpendicular to Surface

We know the field inside the conductor is zero, and the excess charges are all on the surface. The charges produce an electric field outside the conductor. On the surface of conductors in electrostatic equilibrium, the electric field is always perpendicular to the surface. Why? Because if not, charges on the surface of the conductors would move with the electric field.

Charges in Conductors •  Consider a conducting shell, and a negative charge

inside the shell. •  Charges will be “induced” in the conductor to make

the field inside the conductor zero. •  Outside the shell, the field is the same as the field

produced by a charge at the center!

Gauss’ Law: Conducting Plane •  Infinite CONDUCTING plane with

uniform areal charge density s •  E is NORMAL to plane •  Construct Gaussian box as shown. •  Note that E = 0 inside conductor

AEA =0

have, welaw, Gauss' Applyingεσ

0

get wefield, electric for the Solvingεσ=E

Gauss’ Law: Conducting ICPP •  Charged conductor of arbitrary shape: no

symmetry; non-uniform charge density •  What is the electric field near the surface

where the local charge density is σ?

(a) σ/ε0(b) Zero(c) σ/2ε0

AEA =0

have, welaw, Gauss' Applyingεσ

0

get wefield, electric for the Solvingεσ=E

E = 0

+ + +

+ +

+

+

+

+ +

+

+

THIS IS A GENERAL RESULT FOR CONDUCTORS!

23-4 Applying Gauss’ Law: Cylindrical Symmetry

•  23.20 Explain how Gauss’ law is used to derive the electric field magnitude outside a line of charge or a cylindrical surface (such as a plastic rod) with a uniform linear charge density λ.

•  23.21 Apply the relationship between linear charge density λ on a cylindrical surface and the electric field magnitude E at radial distance r from the central axis.

•  23.22 Explain how Gauss’ law can be used to find the electric field magnitude inside a cylindrical non-conducting surface (such as a plastic rod) with a uniform volume charge density ρ.

Learning Objectives

© 2014 John Wiley & Sons, Inc. All rights reserved.

23-4 Applying Gauss’ Law: Cylindrical Symmetry

Figure shows a section of an infinitely long cylindrical plastic rod with a uniform charge density λ. The charge distribution and the field have cylindrical symmetry. To find the field at radius r, we enclose a section of the rod with a concentric Gaussian cylinder of radius r and height h. The net flux through the cylinder from Gauss’ Law reduces to yielding

A Gaussian surface in the form of a closed cylinder surrounds a section of a very long, uniformly charged, cylindrical plastic rod.

© 2014 John Wiley & Sons, Inc. All rights reserved.

Gauss’ Law: Cylindrical Symmetry •  Charge of q = 10 C is uniformly spread over a line of length L = 1 m.

•  Use Gauss’ Law to compute magnitude of E at a perpendicular distance of 1 mm from the center of the line.

• Approximate as infinitely long line — E radiates outwards.

• Choose cylindrical surface of radius r, length L co-axial with line of charge.

r = 1 mm E = ?

L=1 m

Line of Charge: λ = q/L Units: [C/m]

E!"

Gauss’ Law: Cylindrical Symmetry

Φ = EA = E2πrL

00 ελ

εLq ==Φ

E = λ L2πε0rL

= λ2πε0r

= 2k λr

• Approximate as infinitely long line — E radiates outwards.

• Choose cylindrical surface of radius r, length L co-axial with line of charge.

r = 1 mm E = ?

L = 1 m

dA || E so cos θ = 1

λ E!"

E!"

dA! "!!

Acyl = 2πrL

Compare to Finite Line Example From Last Week! Ey = kλ r

dx(r2 + x2 )3/2−L/2

L/2

∫

If the Line Is Infinitely Long (L >> r) … Blah, Blah, Blah…

= 2kλLr 4r2 + L2

= kλ r xr2 x2 + r2

⎡

⎣⎢

⎤

⎦⎥−L/2

L/2

Ey =2kλLr L2

= 2kλr

�

! E

r

With Gauss’s Law We Got Same Answer With Two Lines of Algebra!

•  Add the Vectors!

•  Horrible Integral!

•  Trig Substitution!

r

A-Rod