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Physics II
Jan 202009
Get Fuzzy: Jan 20, 2009
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HITT Quiz
Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters. A third identical conducting sphere C is uncharged. Sphere C is first touched to A, then to B, and finally removed. As a result, the electrostatic force between A and B, which was originally F, becomes:
A. F/2 B. F/4 C. 3F/8 D. F/16 E. 0
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A question Two protons (p1
and p2) are on the x axis, as shown here. The directions of the electric Field at points 1, 2, and 3, respectively, are:
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Another Question Positive charge +Q is
uniformly distributed on the upper half of a rod and negative charge –Q is uniformly distributed on the lower half. What is the direction of the electric field at point P, on the perpendicular bisector of the rod?
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PHY 2049: Physics II Electric fields due to a point charge
(Coulomb’s), a wire and a plane. Charge shells don’t act inside. In an insulator with uniformly
distributed charge, only charge enclosed inside contributes to the field outside.
Gauss’ theorem
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PHY 2049: Physics II
z
Rz
qzkE ˆ2/322
The electric field has been calculated along a line passing through the center of the ring.
It is zero at the center of the ring, increases up to z ~ R and then decreases. It decreases as 1/R3 for large z. Looks like a spring. Problem 76
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Wires, Rings etc..
Inside a uniformly charged ring, E = 0.Inside a uniformly charged spherical shell, E = 0.Electric field lines emerge from a positive charge.They end at a negative charge.Electric field lines do not cross.
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02
E Does not depend on the distance.
C/L 2 dimensionally OK
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PHY 2049: Physics II
E = σ/εo inside
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PHY 2049: Physics II
Flux and Gauss’ theorem
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PHY 2049: Physics II Gaussian
(imaginary) surfaces
Flux = Φ = ∑ E.dA E.dA for a cube is
easy to visualize. Let calculus do it
for a sphere or any other shape.
Φc = qtotal /εo
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S1: Φc = q/ εo S2: Φc = -q/ εo S3: Φc = 0 S4: Φc = 0
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• For a cylinder with axial field, no contribution to flux from the side walls.
• From the far top, >0
• Near (bottom) , <0
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PHY 2049: Physics II No force/field from
the charge outside On inside circle
E.4πr2=q/εo Coulomb’s law
Far outside, E = k 5q/r2
P+
+
+
+ +
+
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PHY 2049: Physics II The electric field
in a metal is zero Charge +q on
inside surface. Because it is
neutral, outside surface must be -q.
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PHY 2049: Physics II
Electric field is radial. E.A = 0 for the top
and bottom surfaces. Sideways:
E 2πrh = λh/ε0
E = λ/ 2πrε0
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PHY 2049: Physics II Plane 2EA = σA/εo
E = σ/2εo
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F+
F-
x-axis
Consider the electric dipole shown in the figure inthe presence of a uniform (constant magnitude and
direction) electric field
Forces and torques exerted on electric dipoles by a uniform electric field
along the -axisThe electric field exerts a force on the positive charge and a force on the negatice charge. The net force on the dipole
0net
E xF qE
F qE
F qE qE
The net torque generated by and about the dipole center is:
sin sin sin sin2 2
In vector form: The electric dipole in a uniform electric field does not move bu
F Fd dF F qEd pE
p E
t can rotate about its center p E 0netF
(22-14)
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A
B
U
180˚
90 90
90
sin
sin cos
U d pE d
U pE d pE p E
Potential energy of an electric dipole in a uniform electric field
pE
E
p
min
At point ( 0) has a minimumvalue It is a position of equilibriums
A
table
UU pE
max
At point ( 180 ) has a maximumvalue It is a position of equilibriumunst le
B
ab
UU pE
U p E
cosU pE
(22-15)
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p
Ef
Fig.b
p
Ei
Fig.a Consider the electric dipole in Fig.a. It has an electricdipole moment and is positioned so that is at an p p
Work done by an external agent to rotate an electricdipole in a uniform electric field
i
f
angle
with respect to a uniform electric field An external agent rotates the electric dipole and bringsit in its final position shown in Fig.b. In this position
is at an angle with respect to
E
p
The work W done by the external agent on the dipoleis equal to the difference between the initial andfinal potential energy of the dipole
cos cos
cos cosf i f i
i f
E
W U U pE pE
W pE
(22-16)
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Summary for today
Electric Fields of a continuous charge distributionField lines
Electric flux and Gauss’ theorem
Dipoles and Torques