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Physics2113
Jonathan Dowling
Lecture 22: WED 14 OCT
Exam 2: Review Session
CH 24–27.3 & HW 05–07
Some links on exam stress:
http://appl003.lsu.edu/slas/cas.nsf/$Content/Stress+Management+Tip+1
http://wso.williams.edu/orgs/peerh/stress/exams.html
http://www.thecalmzone.net/Home/ExamStress.php
http://www.staithes.demon.co.uk/exams.html
Exam 2
• (Ch24) Sec.11 (Electric Potential Energy of a System of Point Charges);
Sec.12 (Potential of Charged Isolated Conductor)
• (Ch 25) Capacitors: capacitance and capacitors; caps in parallel and in series,
dielectrics; energy, field and potential in capacitors.
• (Ch 26) Current and Resistance. Current Density. Ohm’s Law. Power in a
Resistor.
• (Ch 27) Single-Loop Circuits, Series & Parallel Circuits, Multi-loop Circuits.
NO RC Circuits.
• Electric potential:
•
•
•
•
•
– What is the potential produced by a system of charges?
(Several point charges, or a continuous distribution)
Electric field lines, equipotential surfaces: lines go from
+ve to –ve charges; lines are perpendicular to
equipotentials; lines (and equipotentials) never cross each
other…
Electric potential, work and potential energy: work to bring
a charge somewhere is W = –qV (signs!). Potential energy
of a system = negative work done to build it.
Conductors: field and potential inside conductors, and on
the surface.
Shell theorem: systems with spherical symmetry can be
thought of as a single point charge (but how much charge?)
Symmetry, and “infinite” systems.
Electric potential, electric potential energy,
work
In Fig. 25-39, point P is at the center of the rectangle. With V
= 0 at infinity, what is the net electric potential in terms of q/d
at P due to the six charged particles?
Derive an expression in terms of q2/a for the work required to
set up the four-charge configuration of Fig. 25-50, assuming
the charges are initially infinitely far apart.
The electric potential at points in an xy plane is given by V =
(2.0 V/m2)x2 - (4.0 V/m2)y2. What are the magnitude and
direction of the electric field at point (3.0 m, 3.0 m)?
Potential Energy of A System of Charges
• 4 point charges (each +Q) are
connected by strings, forming a
square of side L
• If all four strings suddenly snap,
what is the kinetic energy of each
charge when they are very far
apart?
• Use conservation of energy:
– Final kinetic energy of all four charges
= initial potential energy stored =
energy required to assemble the system
of charges
+Q
+Q
+Q
+Q
Do this from scratch!
Don’t memorize the
formula in the book!
We will change the
numbers!!!
Potential Energy of A System of
Charges: Solution
• No energy needed to bring in
first charge: U1=0
+Q
+Q
+Q
+Q
• Energy needed to bring
2
kQ
in 2nd charge: U = QV =
2
1
L
• Energy needed to bring
in 3rd charge =
kQ 2 kQ 2
U 3 = QV = Q(V1 + V2 ) =
+
L
2L
• Energy needed to bring
in 4th charge =
2kQ 2 kQ 2
U 4 = QV = Q(V1 + V2 + V3 ) =
+
L
2L
Total potential energy is sum of
all the individual terms shown
on left hand side = kQ 2
L
(4 + 2 )
So, final kinetic energy of each
2
charge = kQ
4L
(4 + 2 )
Potential V of Continuous Charge Distributions
r = R¢ 2 + z 2
kdq
dV =
r
V = ò dV
Straight Line Charge: dq= dx
 =Q/L
Curved Line Charge: dq= ds
 =Q/2R
Surface Charge: dq=dA
=Q/R2
dA=2R’dR’
Potential V of Continuous Charge Distributions
Curved Line Charge: dq= ds
 =Q/2R
Straight Line Charge: dq= dx
 =Q/L
Potential V of Continuous Charge Distributions
Surface Charge: dq=dA
=Q/R2
dA=2πR’dR‘
Straight Line Charge: dq= dx
 =bx is given to you.
(d) Potential Voltage:
higher? ✔
lower?
+ work?
(a) E-field does:
– work? ✔
+Vhigh = + + + + + + +
increase? ✔
(c) Potential Energy:
decrease?
+ work? ✔
(b) Work done by me:
– work? -V
low
= -------
ICPP:
Consider a positive and a negative charge, freely moving in a
uniform electric field. True or false?
(a) Positive charge moves to points with lower potential voltage.
(b) Negative charge moves to points with lower potential voltage.
(c) Positive charge moves to a lower potential energy.
(d) Negative charge moves to a lower potential energy.
(a) True
(b) False
(c) True
(d) True
+++++++++
–Q
––––––––
+Q
+V
0
–V
electron
(d) Potential Voltage:
(-)
higher? ✔
lower?
+ work? ✔
(a) E-field does:
– work?
-Vlow = - - - - - - -
-
increase?
(c) Potential Energy:
decrease? ✔
+ work?
(b) Work done by me:
– work? ✔
+Vhigh = + + + + + + +
2
+
+
+
+
downhill
(a) ® ?
¬?
­?
¯?
(b) 1 = + 2 = + 3 = + 4 = - 5 = +
W1 = +10eV
(c) 3 >1 = 2 = 5 > 4
W3 = +20eV
W2 = +10eV
W4 = -10eV
W5 = +10eV
3
D = 2d
For speed set d = e = 1!
æ +e +e ö 3 e
(a) VP = ç + ÷ =
= 3/2
è d 2d ø 2 d
æ +e +e ö 3 e
(b) VP = ç + ÷ =
= 3/ 2
è d 2d ø 2 d
æ +e +e ö 3 e
(c) VP = ç + ÷ =
= 3/2
è d 2d ø 2 d
V
(a)
P
=V
(b)
P
=V
(c)
P
No vectors!
Just add with sign.
One over distance.
Since all charges same and
all distances same all
potentials same.
4
Pa
Va = +
Vb = 0
Vc = -
Pb
Pc
Va > Vc > Vb
dV
DV
Ex =
@
(exact for constant field)
dx
Dx
(a) Since Δx is the same, only |ΔV| matters!
|ΔV1| =200, |ΔV2| =220, |ΔV3| =200
|E2| > |E3| = |E1|
The bigger the voltage drop the stronger the field.
Δx
+
+
+
+
-
+
+
+
+
-
(b) = 3
(c) F = qE = ma
+
+
+
+
-
-
accelerate leftward
DV º V f - Vi
Capacitors
E = /0 = q/A0
E =Vd
q=CV
Cplate = 0A/d
Connected to Battery: V=Constant
Disconnected: Q=Constant
Cplate =  0A/d
Csphere=0ab/(b-a)
C=
•
•
•
Isolated Parallel Plate Capacitor: ICPP
Q e0 A
V
=
d
A parallel plate capacitor of capacitance C is
charged using a battery.
Charge = Q, potential voltage difference = V.
Battery is then disconnected.
If the plate separation is INCREASED, does the
capacitance C:
• Q is fixed!
(a) Increase?
• d increases!
(b) Remain the same?
• C decreases (= 0A/d)
(c) Decrease?
• V=Q/C; V increases.
If the plate separation is INCREASED, does the
Voltage V:
(a) Increase?
(b) Remain the same?
(c) Decrease?
+Q
–Q
Parallel Plate Capacitor & Battery: ICPP
• A parallel plate capacitor of capacitance C is
charged using a battery.
• Charge = Q, potential difference = V.
• Plate separation is INCREASED while battery
remains connected.
• V is fixed constant by battery!
Does the Electric Field Inside: • C decreases (= A/d)
0
(a) Increase?
• Q=CV; Q decreases
• E = σ/0 = Q/0A decreases
(b) Remain the Same?
(c) Decrease?
Battery does work on
capacitor to maintain
constant V!
Q e0 A
C= =
V
d
s
Q
E= =
e0 e0 A
+Q
–Q
Capacitors
Capacitors
Q=CV
In series: same charge
1/Ceq= ∑1/Cj
In parallel: same voltage
Ceq=∑Cj
Capacitors in Series and in
Parallel
• What’s the equivalent capacitance?
• What’s the charge in each capacitor?
• What’s the potential across each capacitor?
• What’s the charge delivered by the battery?
Capacitors: Checkpoints, Questions
•Parallel plates:
Spherical:
•
Cylindrical:
C = 0 A/d
C=
1
1 é1 1ù
- ú
ê
4pe 0 ë a b û
Hooked to battery
V is constant.
Q=VC
C = 20 L/ln(b/a)]
(a) d increases -> C decreases -> Q decreases
(b) a inc -> separation d=b-a dec. -> C inc. -> Q increases
(c) b increases -> separation d=b-a inc.-> C dec. -> Q decreases
PARALLEL:
• V is same for all capacitors
• Total charge = sum of Q
SERIES:
• Q is same for all capacitors
• Total potential difference =
sum of V
(a) Parallel: Voltage is same V on each but charge is q/2 on
each since
q/2+q/2=q.
(b) Series: Charge is same q on each but voltage is V/2 on each
since V/2+V/2=V.
Example: Battery Connected —
Voltage V is Constant but Charge Q Changes
capacitance = C; charge = Q; potential
electric field = E;
•
Initial values:
difference = V;
•
Battery remains connected
•
V is FIXED; Vnew = V (same)
•
Cnew =  C (increases)
Qnew = ( C)V =  Q (increases).
Since Vnew = V, Enew = V/d=E (same)
•
•
dielectric
slab
Energy stored? u=0E2/2 => u= 0E2/2 = E2/2
increases
Example: Battery Disconnected —
Voltage V Changes but Charge Q is Constant
capacitance = C; charge = Q; potential
electric field = E;
•
Initial values:
difference = V;
•
Battery remains disconnected
•
Q is FIXED; Qnew = Q (same)
Cnew =  C (increases)
• Vnew = Q/Cnew = Q/( C) (decreases).
• Since Vnew < V, Enew = Vnew/d = E/ (decreases)
•
Energy stored?
dielectric
slab
u = e0E2 / 2
® ke 0 E
2
new
/ 2 = ke 0 ( E / k ) / 2
= e 0 E 2 / (k 2 ) decreases
2
Current and Resistance
i = dq/dt
Junction rule
R =  L/A
r =  0(1+a(T-T0))
V=iR
E=J
Current and Resistance
A cylindrical resistor of radius 5.0mm and
length 2.0 cm is made of a material that has a
resistivity of 3.5x10-5 m. What are the (a)
current density and (b) the potential difference
when the energy dissipation rate in the resistor
is 1.0W?
Current and Resistance: Checkpoints,
Questions
26.2: Electric Current, Conservation of Charge, and Direction of Current:
Fill in the blanks.
Think water in hose!
2+3= 5
1+ 2 = 3
3+ 2 = 5
2+4 = 6
6+2 = 8
- 3A
¯ 2A
+
+
+
+
-
All quantities defined in terms of + charge movement!
(a) right
(b) right
(c) right
(d) right
DC Circuits
Loop rule
V = iR
P = iV
Single loop
Multiloop
Resistors vs Capacitors
Resistors
Key formula: V=iR
Capacitors
Q=CV
In series: same current
Req=∑Rj
same charge
1/Ceq= ∑1/Cj
In parallel: same voltage
1/Req= ∑1/Rj
same voltage
Ceq=∑Cj
Resistors
in Series and in Parallel
• What’s the equivalent resistance?
• What’s the current in each resistor?
• What’s the potential across each resistor?
• What’s the current delivered by the battery?
• What’s the power dissipated by each resisitor?
Series and Parallel
Series and Parallel
Series and Parallel
Problem: 27.P.018. [406649]
Figure 27-33 shows five 5.00 resistors.
(Hint: For each pair of points, imagine that a battery is connected
across the pair.)
Fig. 27-33
(a) Find the equivalent resistance between points F and H.
(b) Find the equivalent resistance between points F and G.
Slide Rules:
You may bend the wires but not break them.
You may slide any circuit element along a wire so long as you don’t
slide it past a three (or more) point junction or another circuit
element.
Too Many Batteries!
Circuits: Checkpoints, Questions
+E
-iR
-®+
i®
(a) Rightward (EMF is in direction of current)
(b) All tie (no junctions so current is conserved)
(c) b, then a and c tie (Voltage is highest near battery +)
(d) b, then a and c tie (U=qV and assume q is +)
(a) all tie (current is the same in series)
(b) R1 > R2 > R3
Þ iR1 > iR2 > iR3
Þ V1 > V2 > V3
The voltage drop is –iR proportional to R since i is same.
(a) Vbatt < 12V (walking with current voltage drop –ir
(b) Vbatt > 12V (walking against current voltage increase +ir
(c) Vbatt = 12V (no current and so ir=0)