a b c 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)

a

b

c

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.

•Choose either or

•And E constant over surface

• is just the area of the Gaussian surface over which we are integrating.

•Gauss’ Law

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

Gauss’ Law

E dS E dS EdS

0E dS E dS

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

L

R

R

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

• Symmetry E-field of point charge is radially 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!

Gauss Coulomb

• Therefore,

– Gauss’ Law

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

E

+QR

24E dS EdS E dS R E

2

0

4Q

R E

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

qE

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

Gauss’ Law

But,

Thus:

ra

E

a

r

Gauss’ Law and Conductors

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

• But since .0

s

E dS inside 0Q

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

Conducting sphere

++

+

++

+

++

Gauss’ Law and Conductors

• The electric field immediately outside a conductor must be perpendicular to the conductor surface.– Otherwise the charges would move along

the surface until the field was perpendicular everywhere.

• Applying Gauss’ law to a small Gaussian cylinder perpendicular to the surface

• The field just outside a conductor is perpendicular to the surface and proportional to the surface charge density 0 0

0

Q AE dS EA

E

dS

A B

A blue sphere A is contained within a red spherical shell B. There is a charge QA on the blue sphere and charge QB on the red spherical shell.

The electric field in the region between the spheres is completely independent of QB the charge on the red spherical shell.

True

False

Question 1

Question 1

a b

0%0%

1. a

2. b

10

0of5

A B

A blue sphere A is contained within a red spherical shell B. There is a charge QA on the blue sphere and charge QB on the red spherical shell.

The electric field in the region between the spheres is completely independent of QB the charge on the red spherical shell.

True

False

Question 1

Infinite Line of Charge, charge/length=

• 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,

• On the barrel,

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

+ + + + ++ + + +x

y

+ + + + + + + + + + +

Er

h

Er

+ + + + + ++ + +

Charge density on a conducting cylinder• 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

View end on:

Draw Gaussian tube contained within the conducting cylinder

The field within the conducting cylinder is zero

A charge equal and opposite to the line charge is induced on the inner conductor surface that cancels the line charge

b

o

0

0q

E dS q

Charge density on a conducting cylinder

a

b

•Now draw Gaussian tube which surrounds the outer edge

•The tube still contains the line charge and the field is the same as calculated before

o

b

r

02outsideEr

•The charge inside the Gaussian tube also now

=( charge on the outer surface = 02bL)

+ (charge on inner surface + line charge =0 )

Therefore by Gauss’ Law

•A charge equal to the line charge is induced on the outer surface of the cylinder

Question 2 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.

•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|

B) Same as (A) but conducting shell removed

Question 2

a b c

0% 0%0%

1. a

2. b

3. c

10

0of5

Question 2Consider 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|

•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.

• It encloses charge -|Q|, whether or not the uncharged shell is present.

(The field at point X is determined only by the objects with NET CHARGE.)

B) Same as (A) but conducting shell removed

Question 3Consider 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.

E

2

1

-|Q|

•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

Question 3

a b c

0% 0%0%

1. a

2. b

3. c

10

0of5

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

•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

Question 3

• 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|

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

• Chapter 23 of Fishbane• Try Chapter 23 problems 25, 29, 33, 47, 51, 56

Summary• 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