2/6/11. Physics 2020 Lectures and Clicker Quizzes Weeks 34. M. Goldman Spring, 2011. Upcoming exam. Concepts versus facts = thinking versus remembering
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2/6/11
Physics 2020 Lectures and Clicker Quizzes Weeks 34
M. Goldman
Spring, 2011

positive charge, q
++
F = qE
F
+
++
E
E
F
negative charge, q
net charge on surface only
Efield surface
E = 0 inside
A metal in electrostatic equilibrium
The Efield must be perpendicular to the surface (in electrostatic equilibrium), otherwise the component of the Efield along the surface would push electrons along the surface causing movement of charges (and we would not be in equilibrium).
On the surface of metal, if Ex was not zero, there would be a force Fx = q Ex = –e Ex on electrons in the metal pushing them along the surface.
E
Ey
Ex
metal
Put a conducting slab in an electric field, E. Electronsin slab will distribute themselves so that thereis zero net field inside the slab
URBAN LEGEND:
The insulating rubber tires on a car protect you from lightning.
REAL PHYSICS:
The metal body of the car keeps the charges on the outside, away from you! It is essentially a Faraday charge
Remember that Electric Fields inside conductors (metals) are zero when in electrostatic equilibrium.
Inside a hollow metal object (like a car) that has excess charge put on it (from lightning) the charge will be on the outside surface of the object!
However, this is not a case of electrostatic equilibrium. The charges are moving, but by similar principle they stay along the outside conductor and then try to leave via the easiest path (in this case to the ground).
What is the magnitude of the Efield at point P?
A) E = 2kQ / s2
B) E = sqrt(2) kQ / s2
C) E = kQ / ( sqrt(2) s)2
D) zero
E) none of the above
Q
+Q
s
s
s
P
s
Q
+Q
Beware of high voltage
m
Fhand
h
m
m
Choose zeroof P.E. here
Fgravity
Fgravity
Fgravity
P.E. = minimum work done by hand (against gravity) to bring m up a vertical distance, h = Fhandh. Minimum Fhand = Fgravity.
P.E. = (mg)hConservative force independent of path. No work to move m horizontally. Only verticalcomponent of Fhand has to balance gravity.
Electricpotentialenergy =F·h =q·E·h
q
h
E
F
A) It cannot be determined without knowing its mass
B) It is Eqh
C) It is mgh

Electricpotentialenergy =F·h =q·E·h
q
E
h
21
P.E. is negative. Becomes more negative as chargemoves up from h. More negative means decrease!
E.g. 6V < 4V
6 volts
0 volts
ydirection
103 m
V = Eh
E = V/h = 6V/103m
More generally,
6 volts
= 6 kV/m
0 volts
The voltage difference between any two drawn equipotentiallines is the same unless they are labeled otherwise
ydirection
103 m
103 m
103 m
equally spaced
The capacitor field, E, is the same betweenany two equipotential lines. It is uniform in this case.
18 volts
12 volts
6 volts
0 volts
b
Q = 2 C
a
10V
20V
30V
40V
50V
What do equipotential curves look like when E has a different magnitude/direction at each point in space (is spatially nonuniform). Consider point (+)charge, Q:
Red dashed circles are equipotential curves
They are still perpendicular to the electric field lines
The voltage difference is the same between any two equipotential circles
Thespatial separation, s = rlarger  rsmallerbetween circlesshrinks as V increases
5 V
s
s
The electric field is givenmore and more accurately as s decreases by
10 V
15 V
Consider the equipotential lines of a dipole
This is a topo map of the following potential terrain
(+)charges create ‘mountains’, ()charges create ‘valleys.’
Potentials from each charge add as ±numbers to get net V.
Test (+)charges go ‘downhill’, test ( )charges go ‘uphill’.
Crowded lines indicate higher Efield.
The equipotential surfaces around a line of charge (viewed endon) are shown. Each equipotential line is 2 meters from the nearestneighbor equipotential. What is the approximate magnitude of the electric field at point A?
A
1.0V
2.2V
1.4V
1.8V
The equipotential lines are labeled in volts. Spatial distances can be determined from the x and y axes. What is the approximate magnitude and direction of the electric field, E, at point P?
P
Correct answer is: .25 V/m up (negative y direction
P
Voltage is like altitude, and
Efield is like slope
First we find the potential energy, P.E. of another positivecharge, q, put down at the tip of the position vector, r.
P.E. = Work to bring charge q from infinity, where P.E. is defined = 0 to position r. Path doesn't matter for conservative force.
Work NOT EQUAL to force times distance because this force depends on position.
y
q
Q
r
x
Voltage caused by point charge, Q, anywhere on circle of radius r is V = kQ/r
Q
r
r = √(x2+y2)
y
V
y
x
V
y
x
V = kQ/r
r = √(x2+y2)
V
y
x
x
y
V
42
x
r = √(x2+y2+z2)
Equipotential surfaces areconcentric spheres ratherthan circles
z
y
x
E of (+)charge here
() charge
Net E
(+) charge
E of ()charge
Contour map
Relief map
A set of equipotential lines are shown below. Each equipotential line is 2 meters from the nearestneighbor equipotential. What is the approximate electric field at point A?
10V
A
y
8V
6V
x
4V
1 V/m in +x direction
An electron of mass, m, is fired away (to the left) from a large positive charged point starting at a distance r. Its initial velocity (to the left) is v
+Q
e
r
The electron will escape (go out to infinity) if:
A)
B)
C)
D) It can never escape.
Roughguide togrades
A = 8496
A = 80
B+ = 76
B = 72
B = 68
C+ = 64
C = 5660
C = 52
D = 4048
F = 2436

++
+
++
P
dA
dB
d
QB = +2C
QA =  2C
P

++
dA = dB + a
a ≈ d·cosθ
+
++
dB
r
dB
p = Qd = dipole momentr ≈ dB and θ are "polar coordinates"
a
θ
θ
d
QB = +Q
QA =  Q
Q3
1
2
∆A
E = E⊥
+ + + + + + + + + + + + +
σ = ∆Q/∆A =charge per unit area
∆Q
E = E⊥
r
+ + + + + + + + + + + +
Qon sphere
_ _ _ _ _ _ _ _ _ _ _ _
Parallel plate capacitor:
L
Area A = L x W
W
d
We charge the capacitor by (somehow) putting charge onto the plates (i.e. excess electrons or deficit of electrons).
E=0
E = 0
+Q
+++++++++++++++++++++++
Charges are on the inside surfaces of the conductors due to attraction.
d
E

Q
E = 0
E=0
Electric Field (treating plates as infinite) and having area A.
Q and s are assumed to be positive
+Q
+++++++++++++++++++++++
d
E

Q
Define property of capacitorcalled Capacitance:
Q and V are assumed to be positive
For parallel plate capacitor
An electric field of 500 V/m is desired between two parallel plates 10.0 mm apart. How large a Voltage should be applied?
A) 500 Volts
B) 0.5 Volts
C) 2.0 Volts
D) 5.0 Volts
E) None of the above answers
A parallelplate capacitor has square plates of edge length L, separated by a distance d.
If we doubled the dimension L and halve the dimension d, by what factor have we changed the capacitance?
L
L
d