Prof. David R. Jackson Dept. of ECE

Spring 2018

Notes 10 Gauss’s Law I

ECE 3318 Applied Electricity and Magnetism

1

Notes prepared by the EM Group University of Houston

Gauss’s Law

Assume q produces Nf flux lines

qψ =

A charge q is inside a closed surface.

Hence

NS = Nf (all flux lines go through S)

Sf

q NN

ψ

=

2

NS ≡ # flux lines that go through S

ˆS

D n dSψ ≡ ⋅∫

From the picture:

S (closed surface)

ˆ ( )n outward normal

x

y

z

E

q

Gauss’s Law (cont.)

NS = 0

ˆ 0S

D n dSψ = ⋅ =∫

The charge q is now outside the surface

(All flux lines enter and then leave the surface.)

Hence

3

qS

To summarize both cases, we have Gauss’s law:

ˆ enclS

D n dS Q⋅ =∫

By superposition, this law must be true for arbitrary charges.

Gauss’s Law (cont.)

Carl Friedrich Gauss

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n̂ = outward normal

(his signature)

Example

1 2 3

ˆ ˆ ˆ0 2S S S

D n dS q D n dS D n dS q⋅ = ⋅ = ⋅ =∫ ∫ ∫

Note: All of the charges contribute to the electric field in space.

Note: E ≠ 0 on S2 !

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1S2S

3S

2qq−q

Using Gauss’s Law

Gauss’s law can be used to obtain the electric field

for certain problems.

The problems must be highly symmetrical.

⇒ The problem must reduce to one unknown field component (in one of the three coordinate systems).

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Note: When Gauss’s law works, it is usually easier to use than Coulomb’s law

(i.e., the superposition formula).

Choice of Gaussian Surface

Rule 1: S must be a closed surface.

Guideline: Pick S to be ⊥ to E as much as possible

ˆ (LHS)S

D n dS= ⋅∫Left -hand side

ˆS E n D⊥ ⇒

Rule 2: S should go through the observation point (usually called r).

(This simplifies the dot product calculation.)

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ˆ enclS

D n dS Q⋅ =∫

n̂

E

S

Example

Find E

Point charge

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q

x

y

z

Example (cont.)

Assume

ˆ enclS

D n dS Q q⋅ = =∫

Assume

( )r rD D r=

ˆ rD r D=

( )ˆ ˆrS

rS

D r r dS q

D dS q

⋅ =

=

∫

∫

rS

D dS q=∫9

(only a function of r)

(only an r component)

( )24rD r qπ =or Then

ˆ ˆn r=

S

q

r

x

y

z

r

Example (cont.) We then have

( )2

2

4

4

r

r

D r q

qDr

π

π

=

=

20

ˆ4

qE rrπε

=

Hence

10

so

22 C/mˆ

4qD rrπ

=

( )2ˆLHS 4rS

D n dS D rπ= ⋅ =∫

RHS enclQ q= =

11

q

x

y

z

[ ]20

V/mˆ4

qE rrπε

=

Example (cont.)

Summary

Note About Spherical Coordinates Note: In spherical coordinates, the LHS is always:

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( )2ˆ ˆLHS 4r rS S S

D n dS D r dS D dS D rπ= ⋅ = ⋅ = =∫ ∫ ∫

( )ˆ ˆ( )r rD r D r r θ φ= in the direction, and a function of only, not and

( ) ( ), , ( )v rr f r θ φρ θ φ = a function of only, not and

Assumption:

( )2LHS 4rD rπ=

Example

Find E everywhere

Hollow shell of uniform surface charge density

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x

y

z

0s sρ ρ=

a

Example (cont.)

( )24 0r enclD r Qπ = =

V/m0 [ ]E =

Case a) r < a

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Hence

0rD =so

LHS = RHS

x

y

z

0s sρ ρ=

a

a

0sρ

rr

Example (cont.)

( )2 20

20

2

2

4 4

44

4

r encl s

sr

r

D r Q a

aDr

QDr

π ρ π

π ρπ

π

= =

⇒ =

⇒ =

Case b) r > a

20

V/mˆ [ ]4

QE rrπε

=

The electric field outside the sphere of surface charge is the same as from a point charge at the origin.

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Hence

LHS = RHS

ar

0sρ

r

Example (cont.)

20

V/mˆ [ ]4

QE rrπε

=

V/m0 [ ]E =

Note: A similar result holds for the force due to gravity from a

shell of material mass.

Summary

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x

y

z

0s sρ ρ=

a r a>

r a<

Example (cont.)

Note: The electric field is discontinuous as we cross the boundary of a surface

charge density.

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204

Qrπε

x

y

z

0s sρ ρ=

a

ar

rE

( )20/ 4Q aπε

Example

Find E(r) everywhere

Solid sphere of uniform volume charge density

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x

y

z

0v vρ ρ=

a

Example (cont.)

( )2

ˆ

4

enclS

r encl

D n dS Q

D r Qπ

⋅ =

⇒ =

∫

Case a) r < a

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( )encl vV

Q r dVρ= ∫

x

y

z

0v vρ ρ=

a

Gaussian surface S

rr

a0vρ

Example (cont.)

0

0

30

43

encl vV

vV

v

Q dV

dV

r

ρ

ρ

ρ π

=

=

=

∫

∫

Calculate RHS:

20

( )2 30

443r vD r rπ ρ π =

LHS = RHS

Gaussian surface S

rr

a0vρ

Example (cont.)

00

V/mˆ [ ]3vrE r ρε

=

013r vD rρ =

Hence, we have

The vector electric field is then:

21

x

y

z

0v vρ ρ=

a

r a<

r

Example (cont.)

so

30

43encl vQ aρ π =

Case b) r > a

[ ]3

02

0

V/mˆ3

v aE rr

ρε

=

Hence, we have

22

( )2

ˆ

4

enclS

r encl

D n dS Q

D r Qπ

⋅ =

⇒ =

∫

( )2 30

3

0 2

443

3

r v

r v

D r a

aDr

π ρ π

ρ

=

⇒ =

Gaussian surface S

a r

r

0v vρ ρ=

Example (cont.) We can write this as:

The electric field outside the sphere of charge is the same as from a

point charge at the origin.

30

43vQ aρ π =

( )( )

30

20

4 / 3ˆ

3 4 / 3v aE r

rπρ

ε π

=

20

ˆ4

QE rrπε

=

where

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Hence

x

y

z

0v vρ ρ=

a

r a>

r

Example (cont.)

00

V/mˆ [ ]3vrE r ρε

=

[ ]3

02 2

0 0

V/mˆ ˆ3 4

v a QE r rr r

ρε πε

= =

Summary

Note: The electric field is continuous as we

cross the boundary of a volume charge density.

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x

y

z

0v vρ ρ=

a

ra

rE

( )0 0/ 3v aρ ε

r a>

r a<