Gauss law 1

Gauss law

1. 2). Gauss’ Law and Applications
•Coulomb’s Law: force on charge i due to
charge j is
•F
ij is force on i due to presence of j and
acts along line of centres r
ij. If q
i q
j are
same sign then repulsive force is in
direction shown
•Inverse square law of force
( )

ˆ
ˆ
ji
ji
ijjiijjiij
ij2
ij
ji
o
ji3
ji
ji
o
ij
r
r
qq
4
1qq
4
1
rr
rr
rrrrrr
rrr
rr
F
-
-
=-=-=
=-
-
=
pepe
O
r
i
r
j
r
i-r
j
q
i
q
j
F
ij

2. Principle of Superposition
•Total force on one charge i is
•i.e. linear superposition of forces due to all other charges
•Test charge: one which does not influence other ‘real
charges’ – samples the electric field, potential
•Electric field experienced by a test charge q
i ar r
i is
å
¹
=
ij
ij2
ij
j
o
ii
r
q
4
1
q rF ˆ
pe
()å
¹
==
ij
ij2
ij
j
oi
i
ii
r
q
4
1
q
r
F
rE ˆ
pe

3. Electric Field
•Field lines give local direction of field
•Field around positive charge directed
away from charge
•Field around negative charge directed
towards charge
•Principle of superposition used for field
due to a dipole (+ve –ve charge
combination). Which is which?
q
j +ve
q
j -ve

4. Flux of a Vector Field
•Normal component of vector field transports fluid across
element of surface area
•Define surface area element as dS = da
1 x da
2
•Magnitude of normal component of vector field V is
V.dS = |V||dS| cos(Y)
•For current density j
flux through surface S is
Cm
2
s
-1
da
1
da
2
dS
dS = da1 x da2
|dS| = |da
1| |da
2|sin(p/2)
Y
dS`
ò
S surface closed
.dSj

5. •Electric field is vector field (c.f. fluid velocity x density)
•Element of flux of electric field over closed surface E.dS
da
1
da
2
n
q
f
Flux of Electric Field
j
j
jj
ˆ
ˆˆ
ˆ
ˆ
ˆ
θn
naaS
a
θa
x
d dθ sinθ r d x dd
d sinθ rd
dθ rd
2
21
2
1
=
==
=
=
o
oo
2
2
o
q
.d
d
4
q
d dθ sinθ
4
q

1 d dθ sinθr .
r4
q
.d
e
pe
j
pe
j
pe
ò
=
W==
==
S
SE
n.r n
r
SE ˆˆˆ
ˆ
Gauss’ Law Integral Form

6. •Factors of r
2
(area element) and 1/r
2
(inverse square law)
cancel in element of flux E.dS
•E.dS depends only on solid angle dW
da
1
da
2
n
q
f
Integral form of Gauss’ Law
o
i
i
o
21
q
.d
d
4
qq
.d
e
pe
å
ò
=
W
+
=
S
SE
SE
Point charges: q
i enclosed by S
q
1
q
2
v withincharge total)d(
)dv(
.d
V
o
V
=
=
ò
ò
ò
vr
r
SE
r
e
r
S
Charge distribution r(r) enclosed by S

7. Differential form of Gauss’ Law
•Integral form
•Divergence theorem applied to field V, volume v bounded by
surface S
•Divergence theorem applied to electric field E
òòò
Ñ==
V
SS
dv .ddS. VSV. nV
V.n dS
.V dv
o
V
)d(
.d
e
rò
ò
=
rr
SE
S
òò
òò
=Ñ
Ñ=
VV
V
)dv(
1
dv .
dv .d
rE
ESE.
r
e
o
S
oe
r)(
)(.
r
rE=Ñ
Differential form of Gauss’ Law
(Poisson’s Equation)

8. Apply Gauss’ Law to charge sheet
• r (C m
-3
) is the 3D charge density, many applications make use
of the 2D density s (C m
-2
):
•Uniform sheet of charge density s = Q/A
•By symmetry, E is perp. to sheet
•Same everywhere, outwards on both sides
•Surface: cylinder sides + faces
•perp. to sheet, end faces of area dA
•Only end faces contribute to integral
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
E
EdA
ooo
e
s
e
s
e 2
=Þ=Þ=ò
ESE.
S
.dA
E.2dA
Q
d
encl

9. • s’ = Q/2A surface charge density Cm
-2
(c.f. Q/A for sheet)
•E 2dA = s’ dA/e
o
•E = s’/2e
o (outside left surface shown)
Apply Gauss’ Law to charged plate
+


+


+


+


+


+
+


+


+

+


+


+
+


+

+


+


+


+
+


+

+


+


+


+
E
dA
•E = 0 (inside metal plate)
•why??
+




+





+





+
+




+





+





+
•Outside E = s’/2e
o + s’/2e
o = s’/e
o = s/2e
o
•Inside fields from opposite faces cancel

10. Work of moving charge in E field
•F
Coulomb=qE
•Work done on test charge dW
•dW = F
applied.dl = -F
Coulomb.dl = -qE.dl = -qEdl cos q
•dl cos q = dr
•W is independent of the path (E is conservative field)
A
B
q
1
q
r
r
1
r
2
E
dl
q
ò
ò
-=
÷
ø
ö
ç
è
æ
--=
-=
-=
B
A
21o
1
r
r
2
o
1
2
o
1
.dq
r
1
r
1
4
q
q
dr
r
1
4
q
qW
dr
r
1
4
q
qdW
2
1
lE
pe
pe
pe

0=ò
path closed any
lE.d

11. Potential energy function
•Path independence of W leads to potential and potential
energy functions
•Introduce electrostatic potential
•Work done on going from A to B = electrostatic potential
energy difference
•Zero of potential energy is arbitrary
–choose f(r→∞) as zero of energy
r
1
4
q
)(
o
1
pe
f=r
( )
ò
-=
==
B
A
BA
.dq
)(-)(q)PE(-)PE(W
lE
ABAB ff

12. Electrostatic potential
•Work done on test charge moving from A to B when charge q
1
is at the origin
•Change in potential due to charge q
1 a distance of r
B from B
( )
Bo
1
r
2
o
1
B
r
1
4
q
)(
dr
r
1
4
q
.d
-)()(-)(
B
pe
f
pe
fff
=
-=
-=
=¥®
ò
ò
¥
¥
B
lE
BAB


0
( ))(-)(q)PE(-)PE(W
BA
ABAB ff==

13. Electric field from electrostatic potential
•Electric field created by q
1 at r = r
B
•Electric potential created by q
1 at r
B
•Gradient of electric potential
•Electric field is therefore E= – f
3
o
1
r4
qr
E
pe
=
r
1
4
q
r
o
1
B
pe
f=)(
3
o
1
B
r4
q
r
r
pe
f -=Ñ)(

14. Electrostatic energy of charges
In vacuum
•Potential energy of a pair of point charges
•Potential energy of a group of point charges
•Potential energy of a charge distribution
In a dielectric (later)
•Potential energy of free charges

15. Electrostatic energy of point charges
•Work to bring charge q
2 to r
2 from ∞ when q
1 is at r
1 W
2 = q
2 f
2
•NB q
2 f
2 =
q
1 f
1 (Could equally well bring charge q
1 from ∞)
•Work to bring charge q
3 to r
3 from ∞ when q
1 is at r
1 and q
2 is at
r
2 W
3 = q
3 f
3
•Total potential energy of 3 charges = W
2 + W
3
•In general
O
q
1
q
2
r
1 r
2
r
12
12o
1
2
r
1q
pe
j
4
=
O
q
1
q
2
r
1 r
2
r
12
r3
r
13
r
23
23o
2
13o
1
3
r
1q
r
1q
pepe
j
44
+=
åååå
¹<
==
ji jij
j
i
ji jij
j
i
r
q
q
1
2
1
r
q
q
1
W
oo
pepe 44

16. Electrostatic energy of charge
distribution
•For a continuous distribution
òò
ò
ò
-
=
-
=
=
space allspace allo
space allo
space all
)(
d)(d
4
1
2
1
W
)(
d
4
1
)(
)()(d
2
1
W
r'r
r'
r'r r
r'r
r'
r'r
rr r
r
r
pe
r
pe
f
fr

17. Energy in vacuum in terms of E
•Gauss’ law relates r to electric field and potential
•Replace r in energy expression using Gauss’ law
•Expand integrand using identity:
Ñ.yF = yÑ.F + F.Ñy
Exercise: write y = f and F = Ñf to show:
òò
Ñ-==\
Ñ-=Þ-=ÑÞ
-Ñ==Ñ
v
2o
v
2
o
o
2
o
dv
2
dv
2
1
W

and.
ff
e
rf
fer
e
r
f
f
e
r
EE
( )
( )
22
22
.
.
fffff
fffff
Ñ-ÑÑ=ÑÞ
Ñ+Ñ=ÑÑ

18. Energy in vacuum in terms of E
For pair of point charges, contribution of surface term
  1/r   -1/r
2
dA  r
2
overall  -1/r
Let r → ∞ and only the volume term is non-zero
Energy density
( )
( ) ( ) ( )
theorem) e(Divergenc integral volume replaces integral Surface
identity first sGreen' dv.d
2
dvdv .
2
W
v
2o
v
2
v
o
ú
û
ù
ê
ë
é
Ñ-Ñ-=
ú
û
ù
ê
ë
é
Ñ-ÑÑ-=
òò
òò
fff
e
fff
e
S
S
( ) òò
=Ñ=
space all
2o
space all
2o
dvE
2
dv
2
W
e
f
e
)(E
2dv
dW
)(
2o
E
rr
e
r ==