6. gauss’s law (examples) - mcgill physicshilke/142/lecture6.pdf · 2019-01-18 · applying...

6. Gauss’s law (examples)

Announcements:

• Lab schedule is posted. It is important that you read the handout before

the lab and print it out for the lab. There are bonus questions which

require you to prepare accordingly. There is also a lab quiz to be

completed up to 24hrs after the lab.

• Assignment 1 is due Thursday January 24th

Example 1: field from a point charge q

𝐸 = 𝑘0𝑞 Ƹ𝑟

𝑟2=

1

4𝜋𝜀0

𝑞 Ƹ𝑟

𝑟2Coulomb’s law

Example 1: field from a point charge

Gauss’s lawϕ𝐸 = ර

𝜕𝑉

𝐸 ∙ 𝑑𝐴 =𝑄𝑒𝑛𝑐𝑙𝜀0



𝜕𝑉


Electric flux



𝜕𝑉


Closed Gaussian surface

𝜕𝑉 : Surface of sphere



𝜕𝑉


Closed Gaussian surface


𝑑𝐴

Surface element

𝑑𝐴 always points outward

perpendicular to the surface



𝜕𝑉


𝑸𝒆𝒏𝒄𝒍

point charge



𝜕𝑉



𝐸 Ԧ𝑟

Total electric field



𝜕𝑉




𝐸 Ԧ𝑟



𝜕𝑉




𝐸 Ԧ𝑟

𝑑𝐴

Surface element



𝜕𝑉




𝐸 Ԧ𝑟

𝑑𝐴

Surface element

In this configuration:

𝐸 ∥ 𝑑𝐴So

𝐸 ∙ 𝑑𝐴 = E ∗ dA



𝜕𝑉




𝐸 Ԧ𝑟

𝑑𝐴

Surface element

In this configuration:

𝐸 ∥ 𝑑𝐴So

𝐸 ∙ 𝑑𝐴 = E ∗ dA

⇒ ර

𝜕𝑉

𝐸 ∙ 𝑑𝐴 = 𝐸 ර

𝜕𝑉

𝑑𝐴 = 𝐸 ∗ 4𝜋𝑟2



𝜕𝑉




𝐸 Ԧ𝑟

⇒ ර

𝜕𝑉

𝐸 ∙ 𝑑𝐴 = 𝐸 ර

𝜕𝑉

𝑑𝐴 = 𝐸 ∗ 4𝜋𝑟2

= 𝐸 ∗ 4𝜋𝑟2



𝜕𝑉




𝐸 Ԧ𝑟

= 𝐸 ∗ 4𝜋𝑟2

𝐸 =1

4𝜋𝜀0

𝑄𝑒𝑛𝑐𝑙𝑟2



𝜕𝑉




𝐸 Ԧ𝑟

= 𝐸 ∗ 4𝜋𝑟2

𝐸 =1

4𝜋𝜀0

𝑄𝑒𝑛𝑐𝑙𝑟2

𝐸 =1

4𝜋𝜀0

𝑄𝑒𝑛𝑐𝑙 Ƹ𝑟

𝑟2


𝐸 =1

4𝜋𝜀0

𝑞 Ƹ𝑟

𝑟2Coulomb’s law (point charge)

𝐸 =1

4𝜋𝜀0


𝑟2Gauss’s law

Is Coulomb’s law equivalent to Gauss’s law ?

A. Yes

B. No

C. Sometimes

There are equivalent in statics. Coulomb is

a solution to Gauss in statics, but not if the

charges are moving.

𝐸 =1

4𝜋𝜀0න𝑑𝑞

Ƹ𝑟


Need to integrate over a full sphere (a lot of calculus)

Example 2: field from a spherical charge q


𝜕𝑉


Example 2: field from a spherical charge



𝐸 Ԧ𝑟

𝐸 =1

4𝜋𝜀0


𝑟2


𝜕𝑉




𝐸 Ԧ𝑟

𝐸 =1

4𝜋𝜀0


𝑟2



𝜕𝑉




𝐸 Ԧ𝑟

𝐸 =1

4𝜋𝜀0


𝑟2


Field of a spherical charge only

depends on the total charge


𝜕𝑉


Example 3: field outside a hollow spherical charge


𝐸 Ԧ𝑟

𝐸 =1

4𝜋𝜀0


𝑟2


outside


𝜕𝑉


Example 3: field inside a hollow spherical charge


𝐸 Ԧ𝑟

𝐸 = 0

𝑸𝒆𝒏𝒄𝒍=0

inside

𝐸 = 0


𝜕𝑉


Example 4: field inside a filled spherical charge


𝐸 Ԧ𝑟𝑸𝒆𝒏𝒄𝒍

inside

𝐸 =1

4𝜋𝜀0


𝑟2

𝑄𝑒𝑛𝑐𝑙 = 𝑞4𝜋𝑟3

3


𝜕𝑉


Example 4: field inside a filled spherical charge


𝐸 Ԧ𝑟𝑸𝒆𝒏𝒄𝒍

inside

𝐸 =1

4𝜋𝜀0


𝑟2

𝑄𝑒𝑛𝑐𝑙 = 𝑞4𝜋𝑟3

3

𝐸 =𝑞𝑟 Ƹ𝑟

3𝜀0

Inside the sphere, E varies linearly with r

▪ E → 0 as r → 0

The field outside the sphere is equivalent to

that of a point charge located at the center

of the sphere.

Section 24.3

Example 4: field inside and outside a filled spherical charge Q

Example 5: field from an infinite plane surface

𝐸 =1

4𝜋𝜀0න𝑑𝑞

Ƹ𝑟


Need to integrate over a full disk


Coulomb’s law



𝜕𝑉


must be perpendicular to the plane and must have the same magnitude at all points equidistant from the plane.

Choose a small cylinder whose axis is perpendicular to the plane for the gaussian surface.

is parallel to the curved surface and there

is no contribution to the surface area from

this curved part of the cylinder.

The flux through each end of the cylinder is

EA and so the total flux is 2EA.

E

Section 24.3

E



𝜕𝑉


The total charge in the enclosed surface is

𝑄𝑒𝑛𝑐𝑙 =σA.

Applying Gauss’s law:

Note, this does not depend on r.

Therefore, the field is uniform everywhere.

22

E

o o

σA σEA and E

ε ε

Section 24.3



𝜕𝑉


Is E =𝑄𝑒𝑛𝑐𝑙

4𝜋𝑟2𝜀0?

A. Yes

B. No


Gauss: ׯ𝜕𝑉 𝐸 ∙ 𝑑𝐴 =𝑄𝑒𝑛𝑐𝑙

𝜀0

Because E is not aligned with dA nor is E constant along the Gaussian surface

Enter Question Text

A. a

B. b

C. c

D. d

c has the largest flux, a and b the same flux and d zero flux

6. gauss’s law (examples) - mcgill physicshilke/142/lecture6.pdf · 2019-01-18 · applying...

Documents