PM Reyes Notes on Taxation 2 - Valued Added Tax (Working Draft)
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Added Notes
August 31, 2005
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Definitions Electric Field = Force per unit charge. Types of charge distributions
Point Charges Lines of Charge Areas of Charge Volumes of Charge
General:
unitjeschall
jqk rE arg
2r
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Point ChargesIn the Figure, the four particles are fixed in place and have charges q1 = q2
= +5e, q3 = +3e, and q4 = -12e. Distance d = 9.0 mm. What is the
magnitude of the net electric field at point P due to the particles?
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Line of Charge
Need =charge per unit length
unitr
dsk rE
2
ds
dq= ds
r
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A Harder Problem
A line of charge=charge/length
setupsetup
dx
L
r
x
dE dEy
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2/
02/322
2/
02/322
22
2
2
22
)(2
)(2
)()cos(
)(
)cos(
L
x
L
x
L
Lx
xr
dxkrE
xr
dxrkE
xr
r
xr
dxkE
(standard integral)
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Completing the Math
r
kL
r
klE
Lr
L
Lrr
kLE
x
x
2
2
4
:line long VERY a oflimit In the
4
:nintegratio theDoing
22
22
1/r dependence
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Surface Charge
Need Surface Charge density = =charge per unit area. unitr
dAk rE
2
dA
r runit
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The Geometry
Define surface charge density=charge/unit-area
dq=dA
dA=2rdr
(z2+r2)1/2
dq= x dA = 2rdr
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(z2+r2)1/2
R
z
z
rz
rdrzkE
rz
z
rz
drrk
rz
dqkdE
02/322
2/1222222
2
2)cos(
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(z2+r2)1/2
Final Result
0z
220
2E
,R
12
When
Rz
zEz
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Look at the “Field Lines”
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Kinds of continuously distributed charges Line of charge
or sometimes = the charge per unit length. dq=ds (ds= differential of length along the line)
Area = charge per unit area dq=dA dA = dxdy (rectangular coordinates) dA= 2rdr for elemental ring of charge
Volume =charge per unit volume dq=dV dV=dxdydz or 4r2dr or some other expressions we will look at later.
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The Sphere
dqr
thk=dr
dq=dV= x surface area x thickness= x 4r2 x dr
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Summary
222
,2
2
2
)()()(
r
rdsk
r
rdAk
r
rdVk
r
Qk
q
General
r
Qk
q
r
qQk
unitjj
jjj
unit
unit
E
rF
EE
rF
E
rF
(Note: I left off the unit vectors in the lastequation set, but be aware that they should
be there.)
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The figure shows two concentric rings, of radii R and R ' = 3.25R, that lie on the same plane. Point P lies on the central z axis, at distance D = 1.90R from the center of the rings. The smaller ring has uniformly distributed charge +Q. What must be the uniformly distributed charge on the larger ring if the net electric field at point P due to the two rings is to be zero?[-5.39] Q
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The figure shows a plastic ring of radius R = 46.0 cm. Two small charged beads are on the ring: Bead 1 of charge +2.00 µC is fixed in place at the left side; bead 2 of charge +5.60 µC can be moved along the ring. The two beads produce a net electric field of magnitude E at the center of the ring. At what positive and negative values of angle should bead 2 be positioned such that E = 2.00 105 N/C? (Measure the angle from the positive x axis taking counterclockwise to be positive.)
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In the figure, a thin glass rod forms a semicircle of radius r = 4.00 cm. Charge is uniformly distributed along the rod, with +q = 4.00 pC in the upper half and -q = -4.00 pC in the lower half.
(a) What is the magnitude of the electric field at P, the center of the semicircle?[28.6] N/C(b) What is its direction?[-90]° (counterclockwise from the positive x axis)
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d
ds