Gauss Law

a

b

c

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 Lawof 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.

• 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

•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 then

If then

(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.

•With these two conditions we can bring E outside of the integral…and:

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

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

Gauss’ Law…made easy

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

z

R

LR

you may use different E’sfor different surfaces

of your “object”

ab

c

x

y

z

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

+QR

•Why?E normal to every point on the surface

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

•Therefore, !

–Gauss’ Law

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

surface

Uniform charged sphere

• 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

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

a

r

Gauss’

Law

20

1

4

q

r

same as point charge!

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

• Inside sphere: (r < a)– We still have spherical symmetry centered on the center of

the sphere of charge.– Therefore, choose Gaussian surface = sphere of radius r

a

r

Gauss’ Law

But,

Thus:

ra

E

Gauss’ Law and Conductors

• We know that E=0 inside a conductor (otherwise the charges would move).

• But since .0E dS

inside 0Q

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

Conducting sphere

++

+

++

+

++

Lecture 4, ACT 1 Consider the following two topologies:

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

1A •Compare the electric field at point X in cases A and B:

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

E

2

1

-|Q|

1B •What is the surface charge density 1 on the inner surface of the conducting shell in case A?

(a) 1 < (b) 1 = (c) 1 >

B) Same as (A) but conducting shell removed

Lecture 4, ACT 1 Consider the following two topologies:

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

E

2

1

-|Q|

1A •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.)

Consider the following two topologies:

A solid non-conducting sphere carries a total charge Q = -3 C and is surrounded by an uncharged conducting spherical shell.

B) Same as (A) but conducting shell removed

1B•What is the surface charge density 1 on the inner surface of the conducting shell in case A?

(a) 1 < (b) 1 = (c) 1 >

E

2

1

Lecture 4, ACT 1

• 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, 1 must be positive, to cancel the charge -|Q|

• By the way, to calculate the actual value: 1 = -Q / (4 r12)

-|Q|

Infinite Line of Charge• Symmetry E-field

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

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

•Apply Gauss’ Law:• On the ends,

AND• On the barrel,

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

+ + + + ++ + + +x

y

+ + + + + + + + + + +

Er

h

Er

+ + + + + ++ + +

2

Lecture 4, ACT 2• A line charge (C/m) is placed along

the axis of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown.

– What is the value of the charge density o (C/m2) on the outer surface of the cylinder?

(a) (b) (c)

a

b

View end on:

b

Draw Gaussian tube which surrounds only the outer edge

o0

Infinite sheet of charge

• Symmetry:

direction of E = x-axis

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

A

E E

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

• Apply Gauss' Law:

• The charge enclosed = A

Therefore, Gauss’ Law

• On the barrel,

• On the ends,

Two Infinite Sheets(into the screen)

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

– Superposition

– Gaussian surface encloses zero charge

E=0 E=0

E

+

+

++

+

+

+

+

+

+

-

-

--

-

--

-

-

-+

+

---

A

A

0

• Field inside is NOT zero:

– Superposition

– Gaussian surface encloses non-zero charge

• How to do practically all of the homework problems• Gauss’ Law is ALWAYS VALID!

Gauss’ Law: Help for the Problems

• 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).

LHS:

RHS: q = ALL charge inside radius r

LHS:

RHS: q = ALL charge inside radius r, length L

LHS:

RHS: q = ALL charge inside cylinder =A

• Spherical Symmetry: Gaussian surface = Sphere of radius r

• Planar Symmetry: Gaussian surface = Cylinder of area A

• Cylindrical symmetry: Gaussian surface = cylinder of radius r

Sheets of Charge1 σRσL

A B C D

Uncharged Conductor

Hints:1. Assume is positive. It it’s negative, the answer will still work.2. Assume to the right.3. Use superposition, but keep signs straight4. Think about which way a (positive) test charge would move.

ˆ+x

1 1 1A B C D

0 0 0

σ +σ σˆ ˆ ˆE x E x E 0 E x

2E 2E 2E

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

σ1 + 2σ L = 0

σ L = σ R = 1σ

2

1+σ

2

Summary

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

• Gauss’ Law: Electric field flux through a closed surface is proportional to the net charge enclosed– 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

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.

• How is the charge distributed on the sphere?

• How is the charge distributed on the spherical shell?

• What is the electric field at r < R1?

Between R1 and R2? At r > R2?

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

charges?)

R1

R2

Q1

Q2

A

B

C

D

• How is the charge distributed on the sphere?A

* 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

+

+

+

+ +

+

++

• How is the charge distributed on the spherical shell?

B

* 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

22

1inner

R4

Q

and

22

12

2

12outer

R4

Q2

R4

QQ

(C) r < R1:

Inside the conducting sphere

* The electric field inside a conductor is zero.

.0E

(C) Between R1 and R2 : R1 < r < R2

Charge enclosed = Q1

1

2ˆ

QE k r

r

(C) r > R2

Charge enclosed = Q1 + Q2

1 2 1

2 20

1 2ˆ ˆ

4

Q Q QE r k r

r r

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

C

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

charges?)

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.

--

---

-

- -

---

--

- - - - --

-

--

Also, for r < R2 .0E

and for r > R2 r̂r

Q2kE

21

Let’s try some numbersQ1 = 10C R1 = 5cm

Q2 = -30C R2 = 7cm

inner=-162C/m2

outer = -325C/m2

Electric field r < 5cm: Er(r = 4cm) = 0 N/C

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

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

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

B

C

D

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 , and the cylindrical shell has a net surface charge density of total.

outer

R

inner

h

total

outer

h

R

total

inner

•How is the charge distributed on the cylindrical shell?

•What is inner?•What is outer?

•What is the electric field at r<R?

•What is the electric field for r>R?

A

B

C

outer

h

R

total

inner

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 = h•Therefore, the charge on the inner surface of the conducting cylindrical shell is

Qinner = -h The total charge is evenly distributed along the inside surface of the cylinder.

Therefore, the inner surface charge density inner is just Qinner divided by total

area of the cylinder: inner = -h / 2Rh = - / 2R •Notice that the result is independent of h.

A1 What is inner?

outer

h

R

total

inner

•We know that the net charge density on the cylinder is total. The

charge densities on the inner and outer surfaces of the cylindrical shell

have to add up to total. Therefore,

outer =total – inner = total + /(2R).

What is outer? A2

h

Gaussian surface

What is the Electric Field at r<R?

•Whenever we are dealing with electric fields created by symmetric charged surfaces, we must always first chose an appropriate Gaussian surface. In this case, for r <R, the surface surrounding the line charge is

actually a cylinder of radius r.•Using Gauss’ Law, the following equation determines the E-field:

2rhEr = qenclosed /

qenclosed is the charge on the enclosed line charge, which is h, and (2rh) is the area of the barrel of the

Gaussian surface.

The result is:

r

R

h

B

rE

•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: h•Net charge enclosed within Gaussian surface, residing on the cylindrical shell: Q= 2Rh total

•Therefore, net charge enclosed is Q + h•The surface area of the barrel of the Gaussian surface is 2rh•Now we can use Gauss’ Law: 2rh E = (Q + h) / o

•You have all you need to find the Electric field now.

h

R

r

Gaussian surface

total

What is the Electric field for r>R?C

Solve for Er to find

Let’s try some numbers

R = 13 cm h = 168 cm total = 528 C/m2 = 50C/m

A1

A2

B •Electric Field (r<R);

Er(r = 5cm) = 1.798 x 107 N/C

C •Electric Field (r>R);

Er(r = 20cm) = 4.328 x 107 N/C

inner = -61.21 C/m2

outer = 589.2 C/m2

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

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

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

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

A

B

C

1 2

x

y

x2x1

A

Solutions

We can use superposition to find .

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

2

x2

y

x

E1E2

x1

1

Ex(x>x2)

B Ex(x1<x<x2)

1 2

y

xx1 x2

E2

E1

This time the point is located to the left of and to the right

of , therefore;

C Ex(x<x1)

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

1 2

y

xx1 x2

E2E1

Let’s add some numbers...

A

x1= -2m x2= 2m 1= +2C/m2 2= -3C/m2

B

C

E1x= 1.130 x 105 N/C

E2x= -1.695 x 105 N/CEx= -0.565 x 105 N/C

E1x= 1.130 x 105 N/C

E2x= 1.695 x 105 N/CEx= 2.825 x 105 N/C

E1x= -1.130 x 105 N/C

E2x= 1.695 x 105 N/C

Ex= 0.565 x 105 N/C