EEE

1. EEE PPT
• NAME:HARSHIL.R.SHAH
• EN NO:151080106025
• BRANCH:CIVIL
• SEM:2
• TOPIC;OHM’S LAW,GAUSS LAW,
FARADE’S LAW
• SUMBITED BY

2. Ohm’s Law
 Every conversion of energy from one
form to another can be related to this
equation.
 In electric circuits the effect we are
trying to establish is the flow of charge,
or current. The potential difference, or
voltage between two points is the cause
Opposition
Cause
Effect =

3. Ohm’s Law
 Simple analogy: Water in a hose
 Electrons in a copper wire are analogous to
water in a hose.
 Consider the pressure valve as the applied
voltage and the size of the hose as the source
of resistance.
 The absence of pressure in the hose, or voltage
across the wire will result in a system without motion
or reaction.
 A small diameter hose will limit the rate at which
water will flow, just as a small diameter copper wire

4. Ohm’s Law
 Developed in 1827 by Georg Simon Ohm
 For a fixed resistance, the greater the voltage
(or pressure) across a resistor, the more the
current.
The more the resistance for the same voltage,
the less the current.
 Current is proportional to the applied voltage
and inversely proportional to the resistance.

5. Ohm’s Law
Where: I = current (amperes, A)
E = voltage (volts, V)
R = resistance (ohms, Ω)
R
E
I =

6. 4.3 - Plotting Ohm’s Law

7. Plotting Ohm’s Law
•Insert FigInsert Fig
4.84.8

8. 2). Gauss’ Law and Applications
• Coulomb’s Law: force on charge i due to
charge j is
• 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
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
−
−
=−=−=
=−
−
=
πεπε
O
ri
rj
ri-rj
qi
qj
Fij

9. 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 qi ar ri is
∑≠
=
ij
ij2
ij
j
o
ii
r
q
4
1
q rF ˆ
πε
( ) ∑≠
==
ij
ij2
ij
j
oi
i
ii
r
q
4
1
q
r
F
rE ˆ
πε

10. 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?
qj +ve
qj -ve

11. 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(Ψ)
• For current density j
flux through surface S is
Cm2
s-1
da1
da2
dS
dS = da1 x da2
|dS| = |da1| |da2|sin(π/2)
Ψ
dS`
∫ Ssurfaceclosed
.dSj

12. • Electric field is vector field (c.f. fluid velocity x density)
• Element of flux of electric field over closed surface E.dS
da1
da2
n
θ
φ
Flux of Electric Field
ϕ
ϕ
ϕϕ
ˆˆˆ
ˆ
ˆ
ˆ
θn
naaS
a
θa
x
ddθsinθrdxdd
dsinθrd
dθrd
2
21
2
1
=
==
=
=
o
oo
2
2
o
q
.d
d
4
q
ddθsinθ
4
q
1ddθsinθr.
r4
q
.d
ε
πε
ϕ
πε
ϕ
πε
∫ =
Ω==
==
S
SE
n.rn
r
SE ˆˆˆ
ˆ
Gauss’ Law Integral Form

13. • Factors of r2
(area element) and 1/r2
(inverse square law)
cancel in element of flux E.dS
• E.dS depends only on solid angle dΩ
da1
da2
n
θ
φ
Integral form of Gauss’ Law
o
i
i
o
21
q
.d
d
4
qq
.d
ε
πε
∑
∫ =
Ω
+
=
S
SE
SE
Point charges: qi enclosed by S
q1
q2
vwithinchargetotal)d(
)dv(
.d
V
o
V
=
=
∫
∫
∫
vr
r
SE
ρ
ε
ρ
S
Charge distribution ρ(r) enclosed by S

14. 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
∫∫∫ ∇==
V
SS
dv.ddS. VSV.nV
V.n dS .V dv
o
V
)d(
.d
ε
ρ∫
∫ =
rr
SE
S
∫∫
∫∫
=∇
∇=
VV
V
)dv(
1
dv.
dv.d
rE
ESE.
ρ
εo
S
oε
ρ )(
)(.
r
rE =∇
Differential form of Gauss’ Law
(Poisson’s Equation)

15. Apply Gauss’ Law to charge sheet
• ρ (C m-3
) is the 3D charge density, many applications make use
of the 2D density σ (C m-2
):
• Uniform sheet of charge density σ = 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 ε
σ
ε
σ
ε 2
=⇒=⇒=∫ ESE.
S
.dA
E.2dA
Q
d encl

16. • σ’ = Q/2A surface charge density Cm-2
(c.f. Q/A for sheet)
• E 2dA = σ’ dA/εo
• E = σ’/2εo (outside left surface shown)
Apply Gauss’ Law to charged plate
++++++
++++++
++++++
++++++
E
dA
• E = 0 (inside metal plate)
• why??
++++
++++
• Outside E = σ’/2εo + σ’/2εo = σ’/εo = σ/2εo
• Inside fields from opposite faces cancel

17. Work of moving charge in E field
• FCoulomb=qE
• Work done on test charge dW
• dW = Fapplied.dl = -FCoulomb.dl = -qE.dl = -qEdl cos θ
• dl cos θ = dr
A
B
q1
q
r
r1
r2
E
dl
θ
∫
∫
−=






−−=
−=
−=
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
πε
πε
πε
0=∫ pathclosedany
lE.d

18. 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 φ(r→∞) as zero of energy
r
1
4
q
)(
o
1
πε
φ =r
( )
∫−=
==
B
A
BA
.dq
)(-)(q)PE(-)PE(W
lE
ABAB φφ

19. Electrostatic potential
• Work done on test charge moving from A to B when charge q1
is at the origin
• Change in potential due to charge q1 a distance of rB from B
( )
Bo
1
r
2
o
1
B
r
1
4
q
)(
dr
r
1
4
q
.d
-)()(-)(
B
πε
φ
πε
φφφ
=
−=
−=
=∞→
∫
∫
∞
∞
B
lE
BAB 0
( ))(-)(q)PE(-)PE(WBA ABAB φφ==

20. Electric field from electrostatic potential
• Electric field created by q1 at r = rB
• Electric potential created by q1 at rB
• Gradient of electric potential
• Electric field is therefore E= – φ
3
o
1
r4
q r
E
πε
=
r
1
4
q
r
o
1
B
πε
φ =)(
3
o
1
B
r4
q
r
r
πε
φ −=∇ )(

21. Electrostatic energy of point charges
• Work to bring charge q2 to r2 from ∞ when q1 is at r1 W2 = q2 φ2
• NB q2 φ2 =q1 φ1 (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 φ3
• Total potential energy of 3 charges = W2 + W3
• In general
O
q1
q2
r1 r2
r12
12o
1
2
r
1q
πε
ϕ
4
=
O
q1
q2
r1 r2
r12
r3
r13
r23
23o
2
13o
1
3
r
1q
r
1q
πεπε
ϕ
44
+=
∑ ∑∑ ∑ ≠<
==
ji j ij
j
i
ji j ij
j
i
r
q
q
1
2
1
r
q
q
1
W
oo πεπε 44

22. Electrostatic energy of charge
distribution
• For a continuous distribution
∫∫
∫
∫
−
=
−
=
=
spaceallspaceallo
spaceallo
spaceall
)(
d)(d
4
1
2
1
W
)(
d
4
1
)(
)()(d
2
1
W
r'r
r'
r'rr
r'r
r'
r'r
rrr
ρ
ρ
πε
ρ
πε
φ
φρ

23. Faraday's Law
•

24. • Magnetic field around a permanent magnet.
B

25. •Magnetic field around a straight conductor
carrying a steady current I.
•Magnitude of B is directly proportional to the current I value and
inversely proportional to the distance from the conductor.

26. Magnetic flux
∫ ⋅=Φ
S
B SdB

αcos⋅⋅=Φ ∫S
B dsB
[ ]
2
11 mTWb
WbB
⋅=
=Φ

27. Faraday’s Law
md d dx
Blx Bl
dt dt dt
Φ
= =
dx
Blv Bl
dt
= =E
m
Therefore,
d
dt
Φ
=E
• CONCLUSION: to produce emf one should make
ANY change in a magnetic flux with time!
•Consider the loop
shown:

28. LENZ’S Law
•The direction of theThe direction of the
emf induced byemf induced by
changing flux willchanging flux will
produce a currentproduce a current
that generates athat generates a
magnetic fieldmagnetic field
opposing the fluxopposing the flux
change thatchange that
produced it.produced it.

29. Lenz’s Law
•B, H
•Lenz’s Law: emf appears and current flows that creates
a magnetic field that opposes the change – in this case an
increase – hence the negative sign in Faraday’s Law.
•B, H
•N •S

•V-, V+
•Iinduced

30. Faraday’s Law for a Single Loop
dt
d
E
Φ
−=ε=

31. Faraday’s Law for a coil having NFaraday’s Law for a coil having N
turnsturns
dt
d
NE
Φ
−=ε=

32. Lenz's
Law
Claim: Direction of induced current must be so
as to oppose the change; otherwise
conservation of energy would be violated.
• Why???
– If current reinforced the change, then
the change would get bigger and that
would in turn induce a larger current
which would increase the change, etc..
– No perpetual motion machine!
Conclusion: Lenz’s law results from energy
conservation principle.

33. •In 1831 Joseph Henry discovered magnetic induction.
The History of Induction
•Joseph
Henry
•(1797-1878)
•Michael Faraday
•(1791-1867)
• Michael Faraday's ideas about conservation
of energy led him to believe that since an
electric current could cause a magnetic field, a
magnetic field should be able to produce an
electric current. He demonstrated this principle
of induction in 1831.
•So the whole thing started 176 years
ago!

34. THANK YOU