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PHYS113 Electricity and Electromagnetism Semester 2; 2002. Professor B. J. Fraser. 1. Electric Charge. What is charge? 700 BC  Greeks write of effects of rubbing amber ( Electrum ) 1600’s  Gilbert shows electrification is a general phenomenon
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Professor B. J. Fraser
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3 nC
2.0 m
q1
q2
+2 nC
+2 nC
2.0 m
Hints for Problem SolvingExample: Electric Forces in a Plane
i
Solution: Forces in a planeForce on q3
q3
F31
F32
F3
q
q1
q2
Electric Field Region
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Q
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F
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Test Charge
Charged Object
Electric Field Strengthi
6 cm
4 cm
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A
q1 = +12 nC
q2 = 12 nC
Field Due to Point ChargesE = E1 + E2 + E3 + …
Worked Example
Field at A due to q1.
A
+q1
 q2
EA2
Field Due to Point ChargesField at A due to q2
i
^
r
dy
dE
L
y
R
q
O
P
L
r
dE
dQ
Field due to a line segmentl = Q/L
Worked Example
But l = Q/L and thus dQ = l dy
cosq = R/r
r2 =(R2 + y2)
Can you do this integral?
Thus, the field from a line charge is proportional to 1/R and not 1/R2.
^
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k
j
r
R
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i
q
L
x
dx
Field Due to a SurfaceWorked Example
Divide the sheet into an infinite collection of line segments, L, long and, dx, wide
dQ = s dA = s L dx
l = dQ/L = s dx
E = 2kl/r = 2 k s dx/r
But i components sum to 0
Can you do this integral?
Thus, the field from a surface in independent of the distance R!
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e
        
Particles in an Electric FieldE = F/q
F = qE = ma
Worked Example
F = ma = qE
a = qE/m
v2 = u2 + 2as = 2as
v = 2.6 x 108 ms1
q
Fishnets and Flux: The Gaussian Surfacefw = vA
fw = vA cosq
fw = vA
dA
E
Gaussian surface
3. Gauss’ Law= 8.85 x 1012 C2 N1 m2
Radial field lines are always normal to sphere
Example: Coulomb’s Law from Gauss’ Law
E.dA = E dA cos(0º) = E dA
E
+q
r
dA
Gaussian surface
P
Applications of Gauss’ LawdA
h
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dA
dA
r
E.dA=0
E.dA=E dA
Gauss’ Law
Rod Charge Density
Compare this with our previous result.
E varies as 1/R
R
E
Charged Spherical ShellWorked Example
Outside the shell
E.dA = E dA (since E dA)
Inside the shell
No field inside the shell
Faraday Cage!!
Worked Example
Outside the sphere:
As before
R
Inside the Solid SphereInside the sphere
E
r
R
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E = 0
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En
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dA
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Why Doesn’t My Radio Worka charged conductor is:
Proof
In mechanics
W a b= Ua  Ub
Ua > Ub
In electricity
Worked Example
Definition:
1 V = 1 volt = 1 J/C
For a uniform field, d  E
1 eV = (1.6 x 1019 C)(1 V)
= 1.6 x 1019 J
Worked Example
p+
r
Worked ExamplesA very simplistic picture
Worked Example
(a) The force on the electron.
(b) The work done on it by the Efield.
(c) Its potential difference from start to finish.
(d) Its change in potential energy.
(e) Its final speed.
(a) Force is in opposite direction to the Efield, magnitude:
(b) Work done by force:
(c) Potential difference is defined as work/unit charge:
Alternatively (e opposite to p+):
Worked Example
DK.E. = 600(1.6 x 1019) = 9.6 x 1017 J
Worked Example
V = Axy2  Byz
where A and B are constants. Find the electric field.
0V
0.25V
0.1V
0.2V
x =0
x =0.5m
Parallel PlatesWorked Example
Worked Example
R
(y2 + x2)½
y
x
P
dy
Lightning conductors
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Accelerators (1929)
(Giancoli Section 44.2, p1115)
Precipitators (See Figure shown)
Photocopiers (1940)
(Giancoli Example 21.5, p555)