Transcript Gauss` Law and Applications

2). Gauss’ Law and Applications
• Coulomb’s Law: force on charge i due to
charge j is
qiq j
1
1 qiq j ˆ
r  rj  
Fij 
r
3 i
2 ij
4o r  r
4o rij
i
j
rij  ri  rj
rij  ri  rj
rˆij 
ri  rj
ri  rj
• Fij is force on i due to presence of j and
acts along line of centres rij. If qi qj are
same sign then repulsive force is in
ri
direction shown
• Inverse square law of force
O
Fij
qi
ri-rj
rj
qj
Principle of Superposition
• Total force on one charge i is
Fi  qi 
j i
1
qj
4o rij
rˆ
2 ij
• 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 qi ar ri is
Fi
1 qj ˆ
Ei ri    
r
2 ij
qi ji 4o rij
Electric Field
• Field lines give local direction of field
qj +ve
• 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?
qj -ve
Flux of a Vector Field
• Normal component of vector field transports fluid across
element of surface area
• Define surface area element as dS = da1 x da2
• Magnitude of normal component of vector field V is
V.dS = |V||dS| cos(Y)
da2
• For current density j
flux through surface S is
 j.dS Cm2s-1
closed surf aceS
dS
da1
Y
dS`
dS = da1 x da2
|dS| = |da1| |da2|sin(/2)
Flux of Electric Field
• Electric field is vector field (c.f. fluid velocity x density)
• Element of flux of electric field over closed surface E.dS
da1  r dθ θˆ
n
da 2  r sinθ d ˆ
da2
q
f
da1
dS  da1 x da 2  r 2 sinθ dθ d nˆ
nˆ  θˆ x ˆ
q rˆ 2
E.d S 
. r sinθ dθ d nˆ rˆ.nˆ  1
2
4o r

q
4o
q
 E.dS  
S
o
sinθ dθ d 
q
4o
d
Gauss’ Law Integral Form
Integral form of Gauss’ Law
• Factors of r2 (area element) and 1/r2 (inverse square law)
cancel in element of flux E.dS
q1  q2
E.d S 
d
• E.dS depends only on solid angle d
4
o
 E.dS 
n
da2
da1
q
q1
f
q2
S
q
i
i
o
Point charges: qi enclosed by S
  (r )dv
V
E
.d
S


S
o
  (r )dv  total charge within v
V
Charge distribution (r) enclosed by S
Differential form of Gauss’ Law
  (r )dr
V
E
.d
S


• Integral form
S
o
• Divergence theorem applied to field V, volume v bounded by
surface S  V.n dS   V.dS   .V dv
S
S
V
V.n dS
.V dv
• Divergence theorem applied to electric field E
 E.dS   .E dv
V
S
1
 .E dv   
V
o
V
 (r )dv
 (r )
.E(r ) 
o
Differential form of Gauss’ Law
(Poisson’s Equation)
Apply Gauss’ Law to charge sheet
•
 (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
dA
By symmetry, E is perp. to sheet
E
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.dS 
S
Qencl
o
 E.2dA 
s .dA
s
E 
o
2 o
E
Apply Gauss’ Law to charged plate
s’ = Q/2A surface charge density Cm-2 (c.f. Q/A for sheet)
E 2dA = s’ dA/o
E = s’/2o (outside left surface shown)
E = 0 (inside metal plate)
why??
E
• Outside E = s’/2o + s’/2o = s’/o = s/2o
• Inside fields from opposite faces cancel
+
+
+
+
+
+
+
+
dA
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
•
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Work of moving charge in E field
•
•
•
•
FCoulomb=qE
Work done on test charge dW
dW = Fapplied.dl = -FCoulomb.dl = -qE.dl = -qEdl cos q
dl cos q = dr
q1 1
dW  q
dr
2
4o r
B
W  q
q1
4o

r2
r1
1 dr
r2
q1  1 1 
 q
  
4o  r1 r2 
B
 q E.d l
A
E
q
q
dl
r2
A
r
r1
q1
 E.dl  0
any closed path
• W is independent of the path (E is conservative field)
Potential energy function
• Path independence of W leads to potential and potential
energy functions
• Introduce electrostatic potential
q1 1
f (r ) 
4o r
• Work done on going from A to B = electrostatic potential
energy difference
WBA  PE(B) - PE(A)  qf (B) - f ( A)
B
• Zero of potential energy is arbitrary
– choose f(r→∞) as zero of energy
 q E.dl
A
Electrostatic potential
• Work done on test charge moving from A to B when charge q1
is at the origin
WBA  PE(B) - PE(A)  qf(B) - f( A)
• Change in potential due to charge q1 a distance of rB from B
f (B) - f ( A   )  f (B) - 0
B
   E.d l


q1
4o
q1

1
f (B) 
4o rB
rB

1
dr
2
r
Electric field from electrostatic potential
q1
• Electric field created by q1 at r = rB
• Electric potential created by q1 at rB
• Gradient of electric potential
• Electric field is therefore E= – f
r
E
4 o r 3
q1 1
f (rB ) 
4 o r
q1
r
f (rB )  
4 o r 3
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
Electrostatic energy of point charges
• Work to bring charge q2 to r2 from ∞ when q1 is at r1 W2 = q2 f2
q1
q1
q2
r12
q2
r12
r13
r23
r1
r2
r1
r2
q1
1
2 
4o r12
O
r3
q1
1
q2 1
3 

4o r13 4o r23
O
• NB q2 f2 = q1 f1 (Could equally well bring charge q1 from ∞)
• Work to bring charge q3 to r3 from ∞ when q1 is at r1 and q2 is
at r2 W3 = q3 f3
• Total potential energy of 3 charges = W2 + W3
• In general
1
qj
1 1
W
qi  

4o i j
2 4o
j rij
qj
q  r
i
i j
j
ij
Electrostatic energy of charge
distribution
• For a continuous distribution
1
W
dr  (r )f (r )

2 all space
f (r ) 
1
4o
1 1
W
2 4o

dr'
all space
 (r' )
r  r'
 dr  (r ) 
all space
all space
dr'
 (r' )
r  r'
Energy in vacuum in terms of E
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•
Gauss’ law relates  to electric field and potential
Replace  in energy expression using Gauss’ law

.E 
and E  f
o

  2f       o 2f
o
o
2
1
W 
f

dv


f

 f dv
2
v
•
2
v
Expand integrand using identity:
.F = .F + F.
Exercise: write  = f and F = f to show:
2
.ff  f 2f  f 
 f 2f  .ff  f 
2
Energy in vacuum in terms of E
o 

W     .ff dv   f  dv 
2 v
v

2
o 

    ff .d S   f  dv  Green' s first identity 
2 S
v

Surface integral replaces volume integral (Divergenc e theorem)
2
For pair of point charges, contribution of surface term
  1/r   -1/r2 dA  r2 overall  -1/r
Let r → ∞ and only the volume term is non-zero
W
o
2
 f  dv 
2
all space
Energy density
o
2
E (r ) 
2
E
 dv
all space
dW  o 2
 E (r )
dv
2