Gauss Law

c

a

b

  ε 0 ∫ E • dS = qenclosed

Today… • Gauss’ Law: Motivation & Definition • Coulomb’s Law as a consequence of Gauss’ Law • Charges on Conductors: – Where are they?

• Applications of Gauss’ Law – – – – –

Uniform Charged Sphere Infinite Line of Charge Infinite Sheet of Charge Two infinite sheets of charge Shortcuts

Appendix: Three Gauss’ Laws examples Text Reference: Chapter 23.2 →23.5 Examples: 23.4,5,6,7,8 and 9

Fundamental Law of Electrostatics • Coulomb’s Law Force between two point charges

OR • Gauss’ Law Relationship between Electric Fields and charges

Gauss’ Law • Gauss’ Law (a FUNDAMENTAL LAW): The net electric flux through any closed surface is proportional to the charge enclosed by that surface.

  qenclosed ™ E  dS   E  0 • How do we use this equation?? •The above equation is ALWAYS TRUE but it doesn’t look easy to use. •It is very useful in finding E when the physical situation exhibits massive SYMMETRY.

Gauss’ Law…made easy    E  ˜ E  dS  qenclosed /  0 •To solve the above equation for E, you have to be able to CHOOSE A CLOSED SURFACE such that the integral is TRIVIAL. (1) Direction: surface must be chosen such that E is known to be either parallel or perpendicular to each piece of the surface; If If

  E || dS   E⊥dS

then then

  E  dS  E dS

  E • dS = 0

(2) 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.

Gauss’ Law…made easy    E  ˜ E  dS  qenclosed /  0 •With these two conditions we can bring E outside of the integral… and:  

∫ E • dS = ∫ EdS = E ∫ dS

Note that ∫ dS is just the area of the Gaussian surface over which we are integrating. Gauss’ Law now takes the form:

E ∫ dS =

q enclosed

ε0

This equation can now be solved for E (at the surface) if we know qenclosed (or for qenclosed if we know E).

Geometry and Surface Integrals • If E is constant over a surface, and normal to it everywhere, we can take E outside the integral, leaving only a surface area   ∫ E • dS = E∫ dS you may use different E’s for different surfaces of your “object”

z c

y

b

a

∫ dS = 2ac + 2bc + 2ab z

x

R R

∫ dS = 4πR

2

L

2 dS = 2 π R + 2πRL ∫

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 radial and •

spherically symmetric Draw a sphere of radius R centered on the charge.

E R +Q

•Why? E normal  to every point on the surface ⇒ E • dS = EdS E has same value at every point on the surface ⇒ can take E outside of the integral!   •Therefore,∫ E • dS = ∫ EdS = E ∫ dS = 4πR 2 E ! –Gauss’ Law

ε0 4πR E =Q 2

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

E=

1 Q 4πε 0 R 2

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

r

a ρ

• Outside sphere: (r>a) – We have spherical symmetry centered on the center of the sphere of charge – Therefore, choose Gaussian surface = hollow sphere of radius r   q 2 E • d S = 4 π r E = ∫ ε0 4 q = πa 3 ρ 3

⇒

Gauss’ Law

ρa 3 E= 3ε0 r 2



1 q 4  0 r 2

same as point charge!

Uniform charged sphere • Outside sphere: (r > a)

ρa 3 E= 3ε0 r 2

a

• Inside sphere: (r < a)

r

ρ

– We still have spherical symmetry centered on the center of the sphere of charge. – Therefore, choose Gaussian surface = sphere of radius r Gauss’ Law But, Thus:

  q 2 E • d S = 4 π r E = ∫ ε0 4 q = π r3 ρ 3 E=

E

ρ r 3ε 0

a

r

Gauss’ Law and Conductors • We know that E=0 inside a conductor (otherwise the charges would move). • But since

   E  dS  0

 Qinside  0

Charges on a conductor only reside on the surface(s)! + + +

+

+ +

+ + Conducting sphere

.

Lecture 4, ACT 1 Consider the following two topologies: A)

B)

A solid non-conducting sphere carries a total charge Q = -3 µC distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell.

σ2 σ1 -|Q|

E

Same as (A) but conducting shell removed

1A

•Compare the electric field at point X in cases A and B: (a) EA < EB

1B

(b) EA = EB

(c) EA > EB

•What is the surface charge density σ 1 on the inner surface of the conducting shell in case A? (b) σ 1 = 0 (c) σ 1 > 0 (a) σ 1 < 0

Lecture 4, ACT 1 Consider the following two topologies: •

A solid non-conducting sphere carries a total charge Q = -3 µC distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell.

1A

σ2 σ1 -|Q|

E

•Compare the electric field at point X in cases A and B: (a) EA < EB

(b) EA = EB

(c) EA > EB

• Select a sphere passing through the point X as the Gaussian surface. •How much charge does it enclose? •Answer: -|Q|, whether or not the uncharged shell is present. (The field at point X is determined only by the objects with NET CHARGE.)

Lecture 4, ACT 1 σ2

Consider the following two topologies: A solid non-conducting sphere carries a total charge Q = -3 µC and is surrounded by an uncharged conducting spherical shell.

σ1 -|Q|

E

B) Same as (A) but conducting shell removed

1B

•What is the surface charge density σ 1 on the inner surface of the conducting shell in case A? (a) σ 1 < 0

• •

•

(b) σ 1 = 0

(c) σ 1 > 0

Inside the conductor, we know the field E = 0 Select a Gaussian surface inside the conductor • Since E = 0 on this surface, the total enclosed charge must be 0 • Therefore, σ 1 must be positive, to cancel the charge -|Q| By the way, to calculate the actual value: σ 1 = -Q / (4 π r12)

Infinite Line of Charge • Symmetry ⇒ E-field must be ⊥ to line and can only depend on distance from line

y

2

Er Er

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

•Apply Gauss’ Law:

  • On the ends, E • dS = 0

  • On the barrel, E • dS = 2πrhE AND q = λ h ∫

⇒

E=

λ 2πε0 r

NOTE: we have obtained here the same result as we did last lecture using Coulomb’s Law. The symmetry makes today’s derivation easier.

Lecture 4, ACT 2 • A line charge λ (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown.

σ 0= ? b a

λ

– What is the value of the charge density σ o (C/m2) on the outer surface of the cylinder? λ λ σ = + (c) o (b) σ o = 0 (a) σ o = − 2πb 2πb View end on: Draw Gaussian tube which surrounds only the outer edge λ E = σo outside 2πε 0 r 0 b ∫ EdS = (2πrL ) Econductor +( 2πrL) Eoutside = q = σ o 2πbL ε0 ε0 Eoutside =

σ b λ λ = o ⇒σo = 2πε 0 r ε 0 r 2πb

Infinite sheet of charge • Symmetry: +σ

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

A

• Apply Gauss' Law: • On the barrel,

  E • dS = 0

x E

E

  ∫ E • dS = 2 AE

• On the ends, • The charge enclosed =

σ A Therefore, Gauss’ Law ⇒ ε 0 ( 2 EA) = σA

σ E= 2ε0

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

Two Infinite Sheets (into the screen)

• Field outside must be zero. Two ways to see: – Superposition – Gaussian surface encloses zero charge

• Field inside is NOT zero: – Superposition – Gaussian surface encloses

non-zero charge

Q = σA 0   ∫ E • dS = AEoutside + AEinside

+ σ E=0 + + + + A + + + + + A + +

σ E= ε0

σ E=0 E

•

Gauss’ Law: Help for the HowProblems to do practically all of the homework problems • Gauss’ Law is ALWAYS VALID!

  ε 0 ∫ E • dS = qenclosed

• What Can You Do With This? If you have (a) spherical, (b) cylindrical, or (c) planar symmetry AND: • If you know the charge (RHS), you can calculate the electric field (LHS) • If you know the field (LHS, usually because E=0 inside conductor), you can calculate the charge (RHS).

• Spherical Symmetry: Gaussian surface = Sphere of radius r   1 q LHS: ε0 ∫ E • dS = 4πε0 r 2 E E= 4πε 0 r 2 RHS: q = ALL charge inside radius r • Cylindrical symmetry: Gaussian surface = cylinder of radius r  λ LHS: ε0 ∫ E • dS = ε0 2πrLE

E=

2πε 0 r RHS: q = ALL charge inside radius r, length L • Planar Symmetry: Gaussian surface = Cylinder of area A   σ LHS: ε0 ∫ E • dS = ε0 2 AE E= RHS: q = ALL charge inside cylinder =σ A

2ε0

Sheets of Charge σL

1

A

B

σR

C

D

E1 + EL - ER = 0 σ1 + σ L - σ R = 0 σ L + σ R = 0 (uncharged conductor) σ1 + 2σ L = 0

σ1 σ L= 2

+σ1 σ R= 2

Uncharged Conductor

σ1 EA  xˆ 2E 0

Hints: • • • •

+σ1 EB  xˆ 2E 0

EC  0

 σ1 ED  xˆ 2E 0

Assume is positive. It it’s negative, the answer will still work. Assume +xˆ to the right. Use superposition, but keep signs straight Think about which way a (positive) test charge would move.

Summary • Gauss’ Law: Electric field flux through a closed surface is proportional to the net   charge enclosed ε 0 ∫ E • dS = ε 0Φ = qenclosed – Gauss’ Law is exact and always true….

• Gauss’ Law makes solving for E-field easy when the symmetry is sufficient – spherical, cylindrical, planar

• Gauss’ Law proves that electric fields vanish in conductor – extra charges reside on surface

Reading Assignment: Chapter 24.1-> 4 examples: 24.1,2 and 4-6

Example 1: spheres • A solid conducting sphere is concentric with a thin conducting shell, as shown • The inner sphere carries a charge Q1, and the spherical shell carries a charge Q2, such that Q2 = -3Q1.

A

• How is the charge distributed on the sphere?

B

• How is the charge distributed on the spherical shell?

C

• What is the electric field at r < R1? Between R1 and R2? At r > R2?

D

• What happens when you connect the two spheres with a wire? (What are the charges?)

Q2 Q1 R1 R2

A

• How is the charge distributed on the sphere?

* The electric field inside a conductor is zero. (A) By Gauss’s Law, there can be no net charge inside the conductor, and the charge must reside on the outside surface of the sphere

+

+

+ +

+ +

+

+

B

• How is the charge distributed on the spherical shell?

* The electric field inside the conducting shell is zero. (B) There can be no net charge inside the conductor, therefore the inner surface of the shell must carry a net charge of -Q1, and the outer surface must carry the charge +Q1 + Q2, so that the net charge on the shell equals Q2. The charges are distributed uniformly over the inner and outer surfaces of the shell, hence

σ inner

Q1 =− 2 4πR2

and

σ outer

Q2 + Q1 − 2Q1 = = 2 2 4πR2 4πR2

C

• What is the Electric Field at r < R1? Between R1 and R2? At r > R2?

* The electric field inside a conductor is zero. (C) r < R1: Inside the conducting sphere (C) Between R1 and R2 : R1 < r < R2 Charge enclosed = Q1

(C) r > R2 Charge enclosed = Q1 + Q2

 E = 0.

 Q1 E  k 2 rˆ r  1 Q1  Q 2 2Q1 E rˆ   k 2 rˆ 2 4 0 r r

D

-

-

- - - -

• What happens when you connect the two spheres with a wire? (What are the charges?)

-

- - -

-

Also, for r < R2

and for r > R2

After electrostatic equilibrium is reached, there is no charge on the inner sphere, and none on the inner surface of the shell. The charge Q1 + Q2 on the outer surface remains.

 E = 0.

 2Q1 E = − k 2 rˆ r

Let’s try some numbers

B

Q1 = 10µC

R1 = 5cm

Q2 = -30µC

R2 = 7cm

σ inner = -162 µC/m2 σ outer = -325 µC/m2

C

Electric field r < 5cm: Er(r = 4cm) = 0 N/C 5cm < r < 7cm: Er(r = 6cm) = 2.5 x 107 N/C

D

r > 7cm: Er(r = 8cm) = -2.81 x 107 N/C

Electric field r > 7cm: Er(r = 9cm) = -2.22 x 107 N/C

Example 2: Cylinders An infinite line of charge passes directly through the middle of a hollow, charged, CONDUCTING infinite cylindrical shell of radius R. We will focus on a segment of the cylindrical shell of length h. The line charge has a linear charge density λ , and the cylindrical shell has a net surface charge density of σ total.

σ total

λ

R

σ inner σ outer

h

σ outer λ

R

σ inner σ total A

h

•How is the charge distributed on the cylindrical shell? •What is σ inner? •What is σ outer?

B

•What is the electric field at r
C

•What is the electric field for r>R?

What is σ inner?

A1

σ outer

λ

R

σ inner σ total

h

The electric field inside the cylindrical shell is zero. Hence, if we choose as our Gaussian surface a cylinder, which lies inside the cylindrical shell, we know that the net charge enclosed is zero. Therefore, there will be a surface charge density on the inside wall of the cylinder to balance out the charge along the line. •The total charge on the enclosed portion (of length h) of the line charge is: Total line charge enclosed = λ h •Therefore, the charge on the inner surface of the conducting cylindrical shell is

Qinner = -λ h

The total charge is evenly distributed along the inside surface of the cylinder.

σ inner is just Qinner divided by total area of the cylinder: σ inner = -λ h / 2π Rh = -λ / 2π R Therefore, the inner surface charge density

What is σ outer?

A2

σ outer λ

R

σ total

σ inner h

•We know that the net charge density on the cylinder is σ total. The charge densities on the inner and outer surfaces of the cylindrical shell have to add up to σ total. Therefore,

σ outer =σ total – σ inner = σ total +λ /(2π R).

B

What is the Electric Field at r
Gaussian surface

λ

r R h

λ •Whenever we are dealing with electric E = fields created by symmetric charged surfaces, we must always first chose an appropriate Gaussian surface. In this case, for r
The result is:

Er =

λ 2πε0 r

C

What is the Electric field for r>R?

Gaussian surface

λ

R

σ total

r

h

•As usual, we must first chose a Gaussian surface as indicated above. We also need to know the net charge enclosed in our Gaussian surface. The net charge is a sum of the following: •Net charge enclosed on the line: λ h •Net charge enclosed within Gaussian surface, residing on the cylindrical shell: Q= 2πRh σ total •Therefore, net charge enclosed is Q + λ h •The surface area of the barrel of the Gaussian surface is 2πrh •Now we can use Gauss’ Law: 2πrh E = (Q + λ h) / ε o •You have all you need to find the Electric field now. Solve for Er to find

σ R+ λ Er = ε r 2πε r 0 0

Let’s try some numbers R = 13 cm

h = 168 cm

σ total = 528 µC/m2

A1

σ inner = -61.21 µC/m2

A2

σ outer = 589.2 µC/m2

B

•Electric Field (r
C

•Electric Field (r>R); Er(r = 20cm) = 4.328 x 107 N/C

λ = 50µC/m

Example 3: planes Suppose there are infinite planes positioned at x1 and x2. The plane at x1 has a positive surface charge density of σ 1 while the plane at x2 has negative surface charge density of σ 2 . Find:

A

the x-component of the electric field at a point x>x2

B

the x-component of the electric field at x1<x<x2

C

the x-component of the electric field at a point x<x1

σ x1

1

y

σ

2

x2

x

Solutions A

Ex(x>x2)

 We can use superposition to find E.   

E = E1 + E2

The E-field desired is to the right of both sheets. Therefore;

E1 x =

+σ1 2ε0

+σ 2 E2 x = 2ε 0

+σ 1 +σ 2 Ex = 2ε0

y

σ

1

x1

σ

x2

2

E2

x

E1

B

Ex(x1<x<x2)

This time the point is located to the left of σ 2 and to the right of σ 1 , therefore;

−σ 2 E2 x = 2ε 0

+σ1 E1 x = 2ε0

+ σ 1 −σ 2 Ex = 2ε 0

y

σ

1

E2

σ

2

E1 x1

x2

x

Ex(x<x1)

C

When the point is located to the left of both sheets;

σ2 E2 x = − 2ε 0

σ1 E1 x = − 2ε0

−σ 1 −σ 2 Ex = 2ε 0

y

σ E1

1

σ

2

E2 x1

x2

x

Let’s add some numbers... x1= -2m A

x2= 2m

σ 1 = +2µC/m2

E1x= 1.130 x 105 N/C E2x= -1.695 x 105 N/C

B

Ex= -0.565 x 105 N/C

E1x= 1.130 x 105 N/C E2x= 1.695 x 105 N/C

C

σ 2 = -3µC/m2

E1x= -1.130 x 105 N/C E2x= 1.695 x 105 N/C

Ex= 2.825 x 105 N/C

Ex= 0.565 x 105 N/C