Download ELECTRIC FIELD ELECTRIC FLUX GAUSS LAW

ELECTRIC FIELD
ELECTRIC FLUX
GAUSS LAW
Today…..
• More on Electric Field:
– Continuous Charge Distributions
• Electric Flux:
– Definition
– How to think about flux
Summary
Electric Field Lines
Electric Field Patterns
Dipole
Point Charge
Infinite
Line of Charge
Two types of electric charge: opposite charges attract,
like charges repel
1 Q1Q2 ∧
F12 =
2 r12
4πε 0 r12
→
Coulomb’s Law:
Electric Fields
→
Charges respond to electric fields:
Charges produce electric fields:
→
F = qE
q
E=k 2
r
Electric Flux
Flux:
Let’s quantify previous discussion about field-line “counting”
Define: electric flux
ψ
through the closed surface S
ψ = ∫ E • dS
S
“S” is surface
of the box
Electric Flux
ψ = ∫ E • dS
S
•What does this new quantity mean?
• The integral is over a CLOSED SURFACE
→
→
• Since E • dS is a SCALAR product, the electric flux is a SCALAR
quantity
→
• The integration vector dS is normal to the surface and points OUT
→
→
of the surface. E • dS is interpreted as the component of E which is
NORMAL to the SURFACE.
• Therefore, the electric flux through a closed surface is the sum of
the normal components of the electric field all over the surface.
• The sign matters!!
Pay attention to the direction of the normal component as it
penetrates the surface… is it “out of” or “into” the surface?
• “Out of” is “+” “into” is “-”
How to think about flux
• We will be interested in net
flux in or out of a closed
surface like this box
• This is the sum of the flux
through each side of the box
– consider each side separately
• Let E-field point in y-direction
→
– then E and
→
→
→
S
are parallel and
→
E • S = E w2
ELECTRIC FLUX DENSITY
We define electric flux
ψ
D = ε0E
in terms of D
ψ = ∫ D • dS
For example, for an infinite sheet of charge
ρs
E=
an ,
2ε 0
⎛ ρs ⎞
ρs
⎟⎟ a n =
D = ε 0 ⎜⎜
an
2
⎝ 2ε 0 ⎠
For line charge distribution
ρL
ρL
E=
aρ , D =
aρ
2πε 0 ρ
2πρ
For volume charge distribution
ρV dv
ρV dv
E=∫
a
D
=
a
,
R
∫
2
2 R
4πε 0 R
4πR
Example:
Determine D at (4,0,3) if there is a point charge -5π mC at
(4,0,0) and a line charge 3 π mC/m along the y axis.
Let D = DQ +
DL
where DQ and DL are flux densities due to the
point charge and line charge respectively.
Q
Q(r − r )
DQ = ε 0 E =
aR =
2
' 3
4πR
4π r − r
'
Where r – r’ =(4,0,3) –(4,0,0)=(0,0,3).Hence,
− 5π x 10 − 3 (0,0,3 )
mC 2
DQ =
=
−
0
.
138
a
3
z
m
4π 0,0,3
Also,
ρL
DL =
aρ
2πρ
aρ
(
(4,0,3) − (0,0,0) ) (4,0,3)
=
=
(4,0,3) − (0,0,0)
5
3π
(4ax + 3az ) = 0.24ax + 0.18az mC m 2
DL =
2π (5)(5)
Thus,
D = DQ + DL
C
µ
D = 240 a x + 42 a z
m2
GAUSS’S LAW-MAXWELL’S EQUATION
Gauss’s Law states that the total electric flux ψ through any closed
surface is equal to the total charge, Q enclosed by that surface .
ψ = Qenc
That is
ψ = ∫ dψ = ∫ D • dS
S
= Total Charge enclosed , where
Q = ∫ ρ v dv
v
∴ Q = ∫ D • dS = ∫ ρ v dv
S
v
The integration is performed over a closed surface,
i.e. gaussian surface.
By applying Divergence Theorem
∫ D • dS = ∫ ∇ • Ddv
Where,
ρV = ∇ • D
APPLICATION OF GAUSS’S LAW
A. Point Charge
Since D is everywhere normal to gaussian surface, that is
D = Dr ar
From Gauss Law,
ψ = Qenc
Q = ∫ D • dS = Dr ∫ dS = Dr 4πr
Where ∫ D • dS =
2π
∫φ θ
π
2
2
2
r
sin
θ
d
θ
d
φ
=
4
π
r
∫
=0 =0
Q
∴D =
ar
2
4π r
B. Infinite Line of Charge
Suppose the infinite of uniform charge ρ L C m lies along the z-axis.
Determine D at point P?
D = Dρ aρ
Q = ∫ ρ L dl
but , Q = ∫ D • dS = Dρ ∫ dS = Dρ 2πρL
ρL
D=
aρ
2πρ
C. Infinite Sheet of Charge
Consider the infinite sheet of uniform charge is
ρS
C/m2
D = DZ aZ
∫ρ
∫ D • dS
ρ dS = D ( A + A )
ρ A = D (A + A )
Q =
S
dS , Q =
S
Z
S
D =
Z
ρS
2
aZ
ρS
Therefore , E =
=
aZ
ε 0 2ε 0
D
D. Uniformly Charged Sphere
Gaussian surface for a uniformly charged sphere
Q
enc
∫ρ
=
= ρ
v
dv = ρ
2π
v
π
∫
∫
∫
φ
θ
=0
= ρ
r
v
∫
v
dv
r 2 sin θ drd θ d φ
=0 r=0
4
πr
3
3
And
ψ =
∫
D • dS = D
= D
2π
r
π
= D r 4π r
∫
dS
r sin θ d θ d φ
∫
∫
φ
θ
=0
r
2
=0
2
Hence ,
ψ = Qenc gives
4π r
ρv
D r 4π r =
3
Or ..
3
2
r
D = ρ v ar
3
for 0 > r ≤ a
r ≥ a,
For
Q
enc
∫ρ
=
= ρ
v
dv = ρ
2π
v
r
2
dv
sin θ drd θ d φ
=0 r=0
4
πa
3
3
,
While
ψ =
v
a
∫ θ∫ ∫
φ
=0
= ρ
π
∫
v
∫
D • dS
D r 4π r
2
= D r 4π r
4
=
πa 3ρ
3
2
v
Or ..
a3
D =
ρ va r
2
3r
for r ≥ a
D everywhere is given by:
⎧r
a
ρ
⎪3 v r
D=⎨ 3
a
⎪ 2 ρ v ar
⎩ 3r
0<r ≤a
r≥a