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# Gauss' Law 2 - PowerPoint PPT Presentation

Gauss' Law 2. c. a. b. 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.

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Gauss' Law 2

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Gauss' Law 2

c

a

b

• To solve the above equation forE, 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.

### Gauss’ Law

• 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).

you may use different E’s

for different surfaces

z

y

c

b

a

z

x

R

R

L

### 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

R

+Q

• 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!

Þ

Þ

### Gauss Þ Coulomb

• We now illustrate this for the field of a point charge and prove that Gauss’ Law implies Coulomb’s Law.

• Symmetry Þ E-field of point charge is radially and spherically symmetric

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

E

R

+Q

### Gauss Þ Coulomb

• Therefore,

• Gauss’ Law

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

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

r

a

r

Þ

Gauss’

Law

same as point charge!

### 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 radiusr

r

a

r

E

a

r

### 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 radiusr

Gauss’

Law

But,

Thus:

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

+

+

+

+

+

+

+

+

Conducting

sphere

### Gauss’ Law and Conductors

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

• But since  .

### 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

Question 1

A

B

A blue sphere A is contained within a red spherical shell B. There is a chargeQAon the blue sphere and chargeQBon 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

0

of

5

10

### Question 1

• a

• b

Question 1

A

B

A blue sphere A is contained within a red spherical shell B. There is a chargeQAon the blue sphere and chargeQBon 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

y

Er

Er

+ + + + + + + +

+ + + +

+ + + + +

+ + +

x

+ + +

+ + + + + +

h

### Infinite Line of Charge, charge/length=

• Symmetry ÞE-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,

Þ

• 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.

s0=?

b

a

l

so

b

Charge density on a conducting cylinder

• A line charge l (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 so (C/m2) on the outer surface of the cylinder?

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

s0=?

b

a

l

so

r

b

Charge density on a conducting cylinder

• 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

• The charge inside the Gaussian tube also now

• =( charge on the outer surface = 02bL)

• + (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

(a)EA< EB

(b)EA= EB

(c)EA> EB

Consider the following two topologies:

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

s2

s1

-|Q|

E

B) Same as (A) but conducting shell removed

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

0

of

5

10

### Question 2

• a

• b

• c

Question 2

(a) EA< EB

(b) EA= EB

(c) EA> EB

• Consider the following two topologies:

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

s2

s1

-|Q|

E

B) Same as (A) but conducting shell removed

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

• 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.)

Question 3

(b) s1= 0

(c) s1> 0

(a) s1< 0

Consider the following two topologies:

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

s2

s1

-|Q|

E

B) Same as (A) but conducting shell removed

• What is the surface charge density s1 on the inner surface of the conducting shell in case A?

0

of

5

10

### Question 3

• a

• b

• c

(b)s1= 0

(c)s1> 0

(a)s1< 0

Question 3

s2

Consider the following two topologies:

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

B) Same as (A) but conducting shell removed

s1

-|Q|

E

• What is the surface charge density s1on the inner surface of the conducting shell in case A?

• 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, s1 must be positive, to cancel the charge -|Q|

• By the way, to calculate the actual value: s1= -Q/ (4 pr12)

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

• 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