Download 8/30/16 1 Continuous Charge Distributions: Electric Field and

8/30/16
Quiz 2 – 8/25/16
A positive test charge qo is released from rest at a
distance r away from a charge of +Q and a distance 2r
away from a charge of +2Q. How will the test charge move
immediately after being released?
qo
+Q
r
A. 
B. 
C. 
D. 
E. 
+2Q
2r
To the left
To the right
It doesn’t move at all
Moves upward
Moves downward
Continuous Charge Distributions:
Electric Field and Electric Flux
Lecture 3
Last Week
Discrete Charges
•  Magnitude of the electric field due to a point
charge:
!
E=
1 q
4πε 0 r 2
E+
P
!
E−
•  Electric field due to a dipole at
point P:
E=
qd
p
=
2πε o z 3 2πε o z 3
r+
z
r-
+
d
q+
Dipole
- q- center
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8/30/16
Today
Continuous Charge Distributions
•  Two approaches to calculating electric field
–  Extension of point charge methods – Coulomb’s Law
! kq
E = 2 rˆ →
r
! k dq
dE = 2 rˆ
r
Focus on this today
•  Integrate
–  Gauss’ Law
!∫
•  Flux (E•A) = enclosed charge
S
En dA =
1
Q
ε 0 inside
Text Reference: Chapter 22.1
Many useful examples – 22-1 to 8, etc.
Electric Field from Coulomb’s Law
•  Discrete charges vs. continuous charge distribution
•  Principles remain the same
–  Coulomb’s Law + Law of Superposition
Only change:
+
+
+
+
-
+
-
∑
→
+
+ +
∫
+
+ +
+
+ +
+
+
-
Electric Field from Coulomb’s Law
Bunch of Charges
! P
+
+
+
-
qi+
-
-
!
E=
ri
Continuous Charge Distribution
! P
r
+
dq
-
!
E=
+
k
q
1
∑ i rˆ
4πε 0 i ri 2 i
Summation over
discrete charges
1 dq
rˆ =
r2
0
∫ 4πε
"!
∫ dE
⎧ ρdV volume charge
⎪
dq = ⎨σ dA surface charge
⎪λ dL
line charge
Charge
⎩
density
Integral over continuous
charge distribution
http://www.falstad.com/vector3de/
Good visualization
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8/30/16
Charge Densities
•  How do we represent the charge Q on an extended
object?
small pieces
total charge
of charge
Q
dq
Line of charge:
λ = charge per
unit length [C/m]
dq = λ dx
Surface of charge:
σ = charge per
unit area [C/m2]
dq = σ dA = σ dx dy
(Cartesian coordinates)
Volume of Charge:
ρ = charge per
unit volume [C/m3]
dq = ρ dV = ρ dx dy dz
(Cartesian coordinates)
Continuous Charge Distributions
•  Terminology used in the text
–  Source point, xs, where charge ΔQ(xs) is located
–  Field point, xp, where field ΔE(xp) is evaluated
–  Vector from xs to xp: r = xp - xs
•  Superposition
–  Add up all the ΔE created by all charge elements ΔQ
•  Limiting case: replace sum by integral over charge
distribution
"!
!
E=
1 dq
rˆ =
r2
0
∫ 4πε
∫ dE
–  Need to rewrite before we are able to evaluate this
–  Some examples…
Charge Distribution Problems
1)  Understand the geometry
2)  Choose dq
3)  Evaluate dE contribution from the infinitesimal
charge element
4)  Exploit symmetry as appropriate
5)  Set up the integral
6)  Solve the integral
7)  The result!
8)  Check limiting cases
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Field Due to Charge on a Line
Calculate the electric field at point P due to a thin rod of
length L of uniform positive charge, with density λ=Q/L
Steps 1-3
!
E = E x iˆ + E y jˆ
•  Pick any infinitesimal segment
of the rod dxs
–  Charge of the segment is then
dq = λdxs
dE x = dEr cosθ
dE y = dEr sinθ
•  Solve first for Ex
k dq
r2
−x
cosθ = s
r
dEr =
dE x =
k λ dxs
r2
x2
dE x = k λ ∫
cosθ
cosθ dxs
r2
x1
Field Due to Charge on a Line
Calculate the electric field at point P due to a thin rod of
length L of uniform positive charge, with density λ=Q/L
Steps 4-7
•  Change integration variable
from xs to θ
tanθ =
yp
sinθ =
dxs = −y p
x2
∫
x1
cosθ dxs
r2
θ2
=
⇒ xs = −y p cot θ
−xs
∫
θ1
yp
r
⇒r=
yp
sinθ
d cot θ
= y p csc 2 θ dθ
dθ
cosθ y p csc 2 θ dθ
=
y p 2 / sin2 θ
1
yp
θ2
∫ cosθ dθ
θ1
Field Due to Charge on a Line
Calculate the electric field at point P due to a thin rod of
length L of uniform positive charge, with density λ=Q/L
Steps 4-7
•  Evaluate integral, solve for Ex
Ex =
kλ
yp
θ2
kλ
θ1
p
∫ cosθ dθ = y (sinθ
2
− sinθ1)
⎛ 1 1⎞
kλ ⎛ y p y p ⎞
⎜ − ⎟ = kλ ⎜ − ⎟
Ex =
y p ⎜⎝ r2 r1 ⎟⎠
⎝ r2 r1 ⎠
r2>0, r1>0
•  Similarly for Ey
⎛ cot θ cot θ ⎞
2
1
E y = −k λ ⎜
−
⎟ y ≠0
r1 ⎠ p
⎝ r2
E y = 0 yp=0
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8/30/16
Field Due to Charge on a Ring
Calculate the electric field along z-axis due to a circular
ring of radius R of uniform positive charge, with density λ
Steps 1-3
•  Pick any infinitesimal segment of the ring
–  Total charge of the segment is then dq = λds
•  The electric field due to this (point)
charge
dE =
1 dq
4πε 0 r 2
•  Symmetry: direction of the field at
point P must be along the positive z
direction
Field Due to Charge on a Ring
Step 4 – Exploit symmetry
•  The z-component of the field:
1 dq
cosθ
4πε 0 r 2
dE z = dE ⋅ cosθ =
1
λ ds
z
⋅
4πε 0 z 2 + R 2 z 2 + R 2
=
(
)(
zλ
=
(
4πε 0 z 2 + R 2
)
3/2
1/2
)
ds
Field Due to Charge on a Ring
Steps 5-7
•  Integrate over the entire ring with λ uniform
–  z and R fixed
E = Ez =
=
∫ dE
z
=
2π R
zλ
(
4πε 0 z 2 + R 2
)
3/2
∫
ds
0
zλ (2π R)
(
4πε 0 z 2 + R 2
E=
)
3/2
qz
(
2
4πε 0 z + R
2
)
3/2
q: the total
charge of the
ring = 2λπR
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8/30/16
Limiting Cases
Step 8
E=
qz
(
4πε 0 z 2 + R 2
)
3/2
•  Two limiting cases:
–  The electric field vanishes as z approaches zero (or at the center
of the ring)
–  The electric field approaches that of a point charge at large
distances (i.e., large z)
E=
qz
(
2
4πε 0 z + R
2
)
3/2
≈
1 q
4πε 0 z 2
Field From Charged Disk
Problem: calculate the electric field along zaxis due to a circular disk of radius R of
uniform positive charge, with density σ.
•  Pick any ring element (of infinitesimal width dr) of
the disk
•  The charge of the element is dq=σ dA=σ (2π r dr)
•  The electric field due to the ring element
dE =
(σ ⋅ 2π rdr )z
(
4πε 0 z 2 + r 2
)
3/2
Field From Charged Disk
•  Integrate over the entire disk
R
E=
0
=
σz
∫ dE = ∫ 4ε
σz
4ε 0
z 2 +R 2
∫
z2
0
2rdr
(z
2
+r
2
)
3/2
=
dX
σz
=
⋅ −2X −1/2
X 3/2 4ε 0
(
σz
4ε 0
)
R
∫
0
(
d z2 + r 2
(z
2
+r
2
)
)
3/2
z 2 +R 2
z2
where we have set X=z2+r2 and dX=2rdr
z>0
z<0
⎞
σ ⎛
z
E=
⎜⎜1−
⎟⎟
2
2
2ε 0 ⎝
z +R ⎠
E =−
⎞
σ ⎛
z
⎜⎜1−
⎟⎟
2
2
2ε 0 ⎝
z +R ⎠
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8/30/16
Limiting Cases z > 0
E=
•  Small distance
z (or large
disk)
⎞
σ ⎛
z
⎜⎜1−
⎟⎟
2ε 0 ⎝
z2 + R 2 ⎠
z → 0 (or R → ∞) , E →
σ
2ε 0
Ez
Discontinuity
at z= 0
z
Limiting Cases z > 0
E=
•  Large
distance z
E=
=
⎞
σ ⎛
z
⎜⎜1−
⎟⎟
2
2
2ε 0 ⎝
z +R ⎠
⎛
⎞
⎟ σ R2
σ ⎜
1
⋅ 2
⎜1−
⎟≈
2
2ε 0 ⎜
1+ R 2 ⎟ 2ε 0 2z
⎝
z ⎠
1 σ (π R 2 )
1 q
=
4πε 0 z 2
4πε 0 z 2
Infinite Sheet of Positive Charge
Ex = −
σ
2ε 0
Uniform electric
fields generated on
both sides of sheet
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
y
Ex =
σ
2ε 0
x
Discontinuity
at x = 0
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8/30/16
y
90° Arc of Charge
x
E
Yʹ
In this coordinate system
you have to deal with both
Ex and Ey
In this coordinate system
you only have to deal with
the horizontal component
of E
E
Xʹ
uniform charge distribution
total charge = Q
90° Arc of Charge (cont’d)
y
rdθ
λ=
θ
x
dE
Q
2π r
=
2Q
πr
4
dq = λ r dθ
k dq
k λ r cosθ dθ
dE x = 2 cosθ =
r
r2
k λ θ2
Ex =
∫ cosθ dθ
r θ1
kλ
=
(sinθ2 − sinθ1)
r
For the 900 arc, θ1 = −450 and θ 2 = 450 , so
Ex =
k λ ⎡ 1 ⎛ 1 ⎞⎤
2k λ
2k 2Q 2 2kQ
− ⎜−
=
=
⎢
⎟⎥ =
r ⎣ 2 ⎝ 2 ⎠⎦
r
r πr
πr 2
8