1

Lecture 3

IndrianaIndriana HidayahHidayahHanungHanung AdiAdi NugrohoNugroho

a

b

c

εo E dS q•∫ =

Calculating Electric Fields

• Coulomb's LawForce between two point chargesCan also be used to calculate E fields

OR• Gauss' Law

Relationship between Electric Fields and chargesUses the concept of Electric flux

2

Lecture 3

Electric DipoleLines of Force

Consider imaginary spheres centered on :

aa) +q (green)

b

b) -q (red)c

c) midpoint (yellow)

• All lines leave a)• All lines enter b)• Equal amounts of leaving and entering lines for c)

Electric Flux• Flux:

Let’s quantify previous discussion about field-line “counting”Define: electric flux ΦΕ through the closed surface S

• What does this new quantity mean?– The integral is an integral over a CLOSED SURFACE– The result (the net electric flux) is a SCALAR quantity– dS is normal to the surface and points OUT

uses the component of E which is NORMAL to the SURFACE– Therefore, the electric flux through a closed surface is the sum

of the normal components of the electric field all over the surface.

– Pay attention to the direction of the normal component as it penetrates the surface…is it “out of” or “into” the surface?

– “Out of is “+” “Into” is “-”

ΦE E dS≡ •∫

SdE •

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Lecture 3

• Imagine a cube of side a positioned in a region of constant electric field, strength E, as shown.– Which of the following statements about

the net electric flux ΦE through the surface of this cube is true?

(a) ΦE = 0 (b) ΦE = 2Ea2 (c) ΦE= 6Ea2

a

a

Gauss' Law• Gauss' Law (a FUNDAMENTAL Law):

The net electric flux through any closed surface is proportional to the charge enclosed by that surface.

• How to Apply??– The above eqn is TRUE always, but it doesn’t look easy to use– It is very useful in finding E when the physical situation exhibits

massive SYMMETRY– To solve the above eqn for E, you have to be able to CHOOSE a

closed surface such that the integral is TRIVIAL» Direction: surface must be chosen such that E is known to be

either parallel or perpendicular to each piece of the surface» Magnitude: surface must be chosen such that E has the same

value at all points on the surface when E is perpendicular to the surface.

» Therefore: that allows you to bring E outside of the integral

enclosedqSdE =Φ=•∫ 00 εε

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Lecture 3

Geometry and Surface IntegralsIf E is constant over a surface, and normal to it everywhere,we can take E outside the integral, leaving only a surface area

abbcacdS 222 ++=∫a b

c

x

y

z

24 RdS π=∫R

z

R

RLRdS ππ 22 2+=∫L

∫∫ =• dSESdE

Gauss ⇒ Coulomb• We now illustrate this for the field of

the point charge and prove that Gauss’Law implies Coulomb’s Law.

• Symmetry ⇒ E field of point charge is radialand spherically symmetric

• Draw a sphere of radius R centered on the charge.

– Why?E normal to every point on surface

E has same value at every point on surface⇒ can take E outside of the integral!

E =

14πε 0

QR2 ε0 4πR2E = Q– Gauss' Law ⇒

E

+QR

We are free to choose the surface in such problems…we call this a “Gaussian” surface

⇒ EdSSdE =•

• Therefore, ERdSEEdSSdE 24π===• ∫ ∫∫ !

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Lecture 3

+ + + + ++ + + +x

y

+ + + + + + + + + + +

Infinite Line of Charge• Symmetry ⇒ E field

must be ⊥ to line and can only depend on distance from line

⇒ Er

=λπε2 0

Er

• Therefore, CHOOSE Gaussian surface to be a cylinder of radius r and length h aligned with the x-axis.

• Apply Gauss' Law:

q h= λAND

h

Er

+ + + + + ++ + +

0=• SdE• On the ends,E dS rhE•∫ = 2π• On the barrel,

Infinite sheet of charge

• Symmetry:direction of E = x-axis

E =σε2 0

Conclusion: An infinite plane sheet of charge creates a CONSTANT electric field .

• Apply Gauss' Law:

• The charge enclosed = σA

( )ε σ0 2EA A=Therefore, Gauss’ Law ⇒

• Therefore, CHOOSE Gaussian surface to be a cylinder whose axis is aligned with the x-axis.

x

E E

+σ

A

0=• SdE• On the barrel,AESdE 2=∫ •• On the ends,

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Lecture 3

Two Infinite Sheets(into screen)

E=0 E=0

AQ σ=E

σ

+

+

++

++

+

σ

+

++

-

-

--

-

--

-

-

-+

+

---

• Field outside the sheets must be zero. Two ways to see:

• Superposition

• Field inside sheets is NOT zero:

• Superposition

• Gaussian surface encloses zero charge A

• Gaussian surface encloses non-zero chg

A

E = σε0

0insideoutside AEAESdE +=•∫

Uniform charged sphere

• Outside sphere: (r > a)• We have spherical symmetry centered on the

center of the sphere of charge• Therefore, choose Gaussian surface = sphere of

radius r

What is the magnitude of the electric field due to a solidsphere of radius a with uniform charge density ρ (C/m3)?

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Lecture 3

Gauss’ Law: Help for the Problems• How to do practically all of the homework problems

• What Can You Do With This??

If you have (a) spherical, (b) cylindrical, or (c) planar symmetry AND:

• If you know the charge you can calculate the electric field

• If you know the field (usually because E=0 inside conductor), you can calculate the charge.

• Gauss’ Law is ALWAYS VALID!! εo E dS q•∫ =

Application of Gauss’ Law:

q = ALL charge inside cylinder=σAE =

σε2 0

• Planar Symmetry: Gaussian surface = Cylinder of area A

q = ALL charge inside radius rE q

r=

14 0

2πε

• Spherical Symmetry: Gaussian surface = Sphere of radius r

q = ALL charge inside radius r, length LE

r=

λπε2 0

• Cylindrical Symmetry: Gaussian surface = Cylinder of radius r

ε πεo 024E dS r E•∫ =

ε π02 rLE=εo E dS•∫

ε εo 02E dS AE•∫ =

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Lecture 3

Insulators vs. Conductors• Insulators – wood, rubber, styrofoam, most

ceramics, etc. • Conductors – copper, gold, exotic ceramics, etc.

• Sometimes just called metals

• Insulators – charges cannot move.– Will usually be evenly spread

throughout object• Conductors – charges free to move.

– on isolated conductors all charges move to surface.

Conductors vs. Insulators

+-

++-

-

-

+

-

-

- -

++- +

+-- -++++++

-

-- -++++++

- +

- + - +-+

++++++- +

- +- +- +

++++++- +

- +- +- +

- +

- + - +-+

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Lecture 3

+++++++

-------

E+++++++

-------

E

Conductors vs. Insulators

++++++

------

+++++++

-------

E+

- ++-

-

- +

- +

+

-+ +

--

+

-+

+

-

-

- +

- +- +- +

- +

- +- +- +

- +

- + - +

-+

- +

- + - +-+

+++++++

-------

E

Ein = 0 Ein < E

Conductors & Insulators

• How do the charges move in a conductor??• Hollow conducting sphere

Charge the inside, all of this charge moves to the outside.

• Consider how charge is carried on macroscopic objects. • We will make the simplifying assumption that there are only two kinds of objects in the world:

• Insulators.. In these materials, once they are charged, the charges ARE NOT FREE TO MOVE. Plastics, glass, and other “bad conductors of electricity” are good examples of insulators.• Conductors.. In these materials, the charges ARE FREE TO MOVE. Metals are good examples of conductors.

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Lecture 3

Charges on a Conductor• Why do the charges always move to the surface of

a conductor ? – Gauss’ Law tells us!!– E = 0 inside a conductor when in equilibrium (electrostatics) !

» Why?If E ≠ 0, then charges would have forces on them and

they would move !

• Therefore from Gauss' Law, the charge on a conductor must only reside on the surface(s) !

Infinite conductingplane

++++++++++++

++++++++++++Conducting

sphere

+++

++

++

+

Consider the following two topologies:

• Compare the electric field at point X in cases A and B:

(a) EA < EB (b) EA = EB (c) EA > EB

E

σ2

A) A solid non-conducting spherecarries charge Q = -3 μC , and is surrounded by an uncharged conducting spherical shell.

B) Same as (A) but conducting shell removed.

σ1

-Q

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Lecture 3

Consider the following two topologies:

(a) σ1 < 0 (b) σ1 = 0 (c) σ1 > 0

• What is the surface charge density σ1 on the inner surface of the conducting shell in case B?

E

σ2

A) A solid non-conducting spherecarries charge Q = -3 μC , and is surrounded by an uncharged conducting spherical shell.

B) Same as (A) but conducting shell removed.

σ1

-Q

Hollow conductors