8.022 (E&M) – Lecture 3
Topics:
„
Electric potential
„
Energy associated with an electric field
„
Gauss’s law in differential form
… and a lot of vector calculus… (yes, again!)
Last time…
What did we learn?
„
„
„
Energy of a system of charges
Electric field
E =
Fq
q
=
U=
1 i=N
∑
2 i =1
j=N
qi q j
j =1
j ≠i
rij
∑
Q
rˆ
| r |2
Gauss’s law in integral form:
Φ=
„
∫
S
E i dA = 4π Qencl
Derived last time, but not rigorously…
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8.022 – Lecture 3
2
1
NB: Gauss’s law only because
E~1/r2. If E ~ anything else,
the r2 would not cancel!!!
Gauss’s law
„
„
„
„
Consider charge in a generic surface S
Surround charge with spherical surface S1
concentric to charge
Consider cone of solid angle dΩ from
charge to surface S through the little sphere
Electric flux through little sphere:
d Φ S1
„
q
= E i dA = ( 2 rˆ )( r 2 d Ω rˆ ) = qd Ω
r
n̂
S
θ
R
Q
r
S1
Electric flux through surface S:
q
R 2d Ω
rˆ i nˆ
rˆ ) i (
nˆ ) = qd Ω
= qd Ω
2
cos θ
cos θ
R
Æ ΦS=ΦS1=4πQ
d Φ S = E i dA = (
„
dΦS=dΦS1
Φ=
∫
S
E i dA = 4π Qencl is valid for ANY shape S.
G. Sciolla – MIT
8.022 – Lecture 3
3
Confirmation of Gauss’s law
Electric field of spherical shell of charges:
⎧Q
rˆ
⎪
E = ⎨r2
⎪⎩ 0
o u t s i d e t h e s h e ll
i n s i d e t h e s h e ll
Can we verify this experimentally?
+
+
+
+
+
+
+
+
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• Charge a spherical surface with
Van de Graaf generator
• Is it charged? (D7 and D8)
• Is Electric Field radial?
Does E~1/r2, eg: φ~1/r?
Neon tube on only when oriented radially
(D24)
• (D29?)
8.022 – Lecture 3
4
2
Confirmation of Gauss’s law (2)
„
„
„
E=0
Cylindrical shell positively charged
Gauus tells us that
„ Einside = 0
„ Eoutside > 0
+
+
+
E>0
Can we verify this experimentally?
„
Demo D26
„ Charge 2 conductive spheres by induction outside the cylinder: one
sphere will be + and the other will be -: it works because Eoutside > 0
„ Try to do the same inside inside cylinder Æ nothing happens because E=0
(explain induction on the board)
G. Sciolla – MIT
8.022 – Lecture 3
5
Energy stored in E:
Squeezing charges…
„ Consider a spherical shell of charge of radius r
„ How much work dW to “squeeze” it to a radius r-dr?
„ Guess the pressure necessary to squeeze it:
y
F QE
Q
=
= E = Eσ
A
A
A
1Q
Q
Eoutside = 2 ; Einside = 0 → Esurface =
2 r2
r
σ
1Q
→ P = Eσ =
σ = 2 ( 4π r 2σ ) = 2πσ 2
2 r2
2r
P=
x
„ We can now calculate dW:
dW = Fdr = ( PA)dr = (2πσ 2 )(4π r 2 )dr = 2πσ 2 dV
(where dV = 4π r 2 dr )
Remembering that E created in dr =4πσ
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⇒
8.022 – Lecture 3
dW =
E2
dV
8π
6
3
Energy stored in the electric field
„
Work done on the system:
„
„
dW =
E2
dV
8π
We do work on the system (dW): same sign charges have been squeezed on a
smaller surface, closer together and they do not like that…
Where does the energy go?
We created electric field where there was none (between r and r-dr)
Æ The electric field we created must be storing the energy
„ Energy is conserved Æ dU = dW
„
„
u=
E2
8π
is the energy density of the electric field E
U =
∫
E2
dV
8π
„
Energy is stored in the E field:
„
NB: integrate over entire space not only where charges are!
„
Entire
space
Example: charged sphere
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8.022 – Lecture 3
Electric potential difference
„
Work to move q from r1 to r2:
2
2
ds
2
W12 = ∫ FI • ds = − ∫ FCoulomb • ds = − q ∫ E ids
1
„
1
1
W12 depends on the test charge q /
Æ define a quantity that is independent of q
and just describes the properties of the space:
φ1 2 ≡
„
„
q
2
W 12
= − ∫ E ids
1
q
Q
r2
r1
Electric potential difference
between P1 and P2
Physical interpretation:
φ12 is work that I must do to move a unit charge from P1 to P2
Units:
„
cgs: statvolts = erg/esu; SI: Volt = N/C; 1 statvolts = “3” 102 V
G. Sciolla – MIT
8.022 – Lecture 3
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4
Electric potential
„
„
The electric potential difference φ12 is defined as the work to move a
unit charge between P1 and P2: we need 2 points!
Can we define similar concept describing the properties of the space?
„
Yes, just fix one of the points (e.g.: P1=infinity):
r
φ ( r ) = − ∫ E ids
∞
„
„
⇐
P otential
Application 1: Calculate φ(r) created by a point charge in the origin:
r
r q
q
φ ( r ) = − ∫ E ids =− ∫ 2 d r =
∞
∞ r
r
Application 2: Calculate potential difference between points P1 and P2:
φ12 = − ∫
r2
r1
E ids =
q
q
− = φ ( P2 ) − φ ( P1 )
r2 r1
Æ Potential difference is really the difference of potentials!
G. Sciolla – MIT
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8.022 – Lecture 3
Potentials of standard charge distributions
q
r
Æ Given this + superposition we can calculate anything!
The potential created by a point charge is φ ( r ) =
„
Potential of N point charges:
φ (r ) =
N
∑
i =1
qi
ri
ρ dV
„
Potential of charges in a volume V:
φ (r ) =
∫
„
Potential of charges on a surface S:
φ (r ) =
∫
„
Potential of charges on a line L:
φ (r ) =
∫
G. Sciolla – MIT
8.022 – Lecture 3
r
V
σ dA
S
L
r
λ dl
r
10
5
Some thoughts on potential
„
Why is potential useful? Isn’t E good enough?
„
„
Potential is a scalar function Æ much easier to integrate than electric
field or force that are vector functions
When is the potential defined?
„
„
Unless you set your reference somehow, the potential has no meaning
Usually we choose φ(infinity)=0
„
„
This does not work always: e.g.: potential created by a line of charges
Careful: do not confuse potential φ(x,y,z) with potential energy of a
system of charges (U)
1 i = N j = N qi q j
„
Potential energy of a system of charges:
U= ∑ ∑
2 i=1 j =1 rij
work done to assemble charge configuration
„
j ≠i
Potential: work to move test charge from infinity to (x,y,z)
φ (r ) =
N
∑
i =1
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8.022 – Lecture 3
qi
ri
11
Energy of electric field revisited
„
Energy stored in a system of charges:
U=
1 i= N
∑
2 i =1
j=N
qi q j
j =1
j ≠i
rij
∑
„
This can be rewritten as follows:
„
q 1
1
q j ∑ i = ∑ q jφ (rj )
∑
2 j ≠i
2 j ≠i
i rij
where φ(rj) is the potential due to all charges excepted for the qj at
the location of qj (rj)
Taking a continuum limit:
U=
U=
1
E2
(
)
ρφ
=
r
dV
∫
∫ 8π dV
2 Volume
Entire
with
charges
space
NB: this works only when φ(infinity)=0
G. Sciolla – MIT
8.022 – Lecture 3
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6
Connection between φ and E
Consider potential difference between a point at r and r+dr:
dφ = − ∫
r+d r
r
E ids ~ − E (r )id r
The infinitesimal change in potential can be written as:
dφ =
⎛ ∂φ ∂φ ∂φ ⎞
∂φ
∂φ
∂φ
dx +
dy +
dz ≡ ⎜
,
,
⎟ • ( d x , dy , d z ) ≡ ∇ φ • dr
∂x
∂y
∂z
⎝ ∂x ∂y ∂z ⎠
E = −∇ φ
Useful info because it allows us to find E given φ
„
Good because φ is much easier to calculate than E
G. Sciolla – MIT
8.022 – Lecture 3
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Getting familiar with gradients…
1d problem:
∇f ( x ) ≡
∂f
xˆ
∂x
The derivative df/dx describes the function’s slope
Æ The gradient describes the change of the function and the
direction of the change
„
2d problem:
∇f ( x , y ) ≡
⎛ ∂f ∂f ⎞
∂f
∂f
xˆ +
yˆ ≡ ⎜ , ⎟
∂x
∂y
⎝ ∂x ∂y ⎠
The interpretation is the same, but in both directions
Æ The gradient points in the direction where the slope is deepest
„
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8.022 – Lecture 3
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7
Visualization of gradients
Given the potential φ(x,y)=sin(x)sin(y), calculate its gradient.
∇φ ( x, y ) = cos( x) sin( y ) xˆ + sin x cos yyˆ
φ (x,y)
grad φ
The gradient
Æ E=-gradφ
points downhill
G. Sciolla – MITalways points uphill
8.022 – Lecture
3
15
Visualization of gradients:
equipotential surfaces
Same potential φ(x,y)=sin(x)sin(y)
φ (x,y)
NB: since equipotential lines are perpendicular to the gradient
Æ equipotential lines are always perpendicular to E
G. Sciolla – MIT
8.022 – Lecture 3
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8
Divergence in E&M (1)
Consider flux of E through surface S:
S2
S2S2new
S20
dA
dA
S
S
S1new new
S10
S
S11
Cut S into 2 surfaces: S1 and S2 with Snew the little surface in between
Φ=
∫
∫
=∫
S1
E idA − ∫
S1
E idA − ∫ E idA = Φ1 + Φ 2
=
S
E idA =
∫
E idA +
S 1− S 1new
S 1new
E idA +
∫
∫
S 2 − S 2 new
E idA − ∫
S2
E idA
S 2 new
E idA
S2
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8.022 – Lecture 3
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Divergence Theorem
„
Let’s continue splitting into smaller volumes
Φ=
i = largeN
∑
Φi =
i =1
„
Æ
i = largeN
∑ ∫
i =1
Si
E idAi =
If we define the divergence of E as
Φ=
largeN
∑ V (∇i E ) → ∫
i =1
Æ
i
V
∫
S
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i = largeN
∑
i =1
Vi
∫
∇i E ≡ lim
V →0
Si
E idAi
Vi
∫
S
E idA
V
∇i EdV
E idA = ∫ ∇i EdV
V
8.022 – Lecture 3
Divergence Theorem
(Gauss’s Theorem)
18
9
Gauss’s law in differential form
Simple application of the divergence theorem:
⎧
∫ S E idA = ∫V ∇i EdV
⎪
⎨
⎪⎩ ∫ S E idA = 4π Q = 4π ∫V ρdV
→
∫
V
(∇i E − 4πρ )dV = 0
This is valid for any surface V:
∇i E = 4πρ
Comments:
„
„
First Maxwell’s equations
Given E, allows to easily extract charge distribution ρ
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8.022 – Lecture 3
19
What’s a divergence?
„
Consider infinitesimal cube centered at P=(x,y,z)
„
Flux of F through the cube in z direction:
z
∆z
∆z
∆z
∆Φ z = ∫ F idA ~∆x∆y[ Fz ( x, y, z + ) − Fz ( x, y, z − )
2
2
top + bottom
„
∆Φ z = (∆x∆y∆z ) lim
∆z →0
„
∆x
∆y
x
Since ∆zÆ0
P
y
∂F
1
∆z
∆z
[ Fz ( x, y, z + ) − Fz ( x, y, z − )] = ∆x∆y∆z z
∆z
2
2
∂z
Similarly for Φx and Φy
∆Φ x = ∆x∆y∆z
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∂Fy
∂Fx
and ∆Φ y = ∆x∆y∆z
∂x
∂y
8.022 – Lecture 3
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10
Divergence in cartesian coordinates
We defined divergence as ∇i F ≡ lim
V →0
∫
S
F idA
V
But what does this really mean?
∇i F ≡ lim
∫
S
∆x → 0
∆y → 0
∆z → 0
F idA
V
∆x∆y∆z (
= lim
∆x →0
∆y →0
∆z →0
=
∂Fx ∂Fy ∂Fz
+
+
)
∂x
∂y
∂z
∆x∆y∆z
∂Fx ∂Fy ∂Fz
+
+
∂x
∂y
∂z
G. Sciolla – MIT
This is the usable expression for the
divergence: easy to calculate!
8.022 – Lecture 3
21
Application of Gauss’s law in
differential form
Problem: given the electric field E(r), calculate the charge
distribution that created it
4
E (r ) = π Krrˆ
3
for r<R
and
E (r )=
4π K 3
R rˆ for r>R
3r 2
Hint: what connects E and ρ? Gauss’s law.
∫
S
E idA = 4π Qencl
∇i E = 4πρ
(integral form)
(differential form)
In cartesian coordinates:
∇•E ≡
⎧ 4π K when r<R
∂Ex ∂E y ∂Ez
+
+
= ... = ⎨
∂x
∂y
∂z
⎩ 0 when r>R
G. Sciolla – MIT
8.022 – Lecture 3
Æ Sphere of radius
R with constant
charge density K
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11
Next time…
„
„
„
Laplace and Poisson equations
Curl and its use in Electrostatics
Into to conductors (?)
G. Sciolla – MIT
8.022 – Lecture 3
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