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Physics2113
Jonathan Dowling
Lecture 20R: MON 13 OCT
Exam 2: Review Session
CH 23.1–9, 24.1–12, 25.1–6,8 26.1–9
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
• (Ch23) Gauss’s Law
• (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, current
density and drift velocity; resistance and resistivity;
Ohm’s law.
• Gauss’ law:  =q/0 . Given the field, what is the
charge enclosed? Given the charges, what is the
flux? Use it to deduce formulas for electric field.
• 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.
Gauss’ law
At each point on the surface of the cube shown in Fig. 24-26,
the electric field is in the z direction. The length of each edge
of the cube is 2.3 m. On the top surface of the cube E = -38 k
N/C, and on the bottom face of the cube E = +11 k N/C.
Determine the net charge contained within the cube.
[-2.29e-09] C
Gauss’s Law: Cylinder, Plane, Sphere
Problem: Gauss’ Law to Find E
Gauss’ law
A long, non conducting, solid cylinder of radius 4.1 cm has a
nonuniform volume charge density that is a function of
the radial distance r from the axis of the cylinder, as given
by r = Ar2, with A = 2.3 µC/m5.
(a)What is the magnitude of the electric field at a radial
distance of 3.1 cm from the axis of the cylinder?
(b)What is the magnitude of the electric field at a radial
distance of 5.1 cm from the axis of the cylinder?
The figure shows conducting plates with area A=1m2, and the
potential on each plate. Assume you are far from the edges of the
plates.
• What is the electric field between the plates in each case?
• What (and where) is the charge density on the plates in case (1)?
• What happens to an electron released midway between the
plates in case (1)?
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.
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
Current and Resistance
i = dq/dt
Junction rule
R =  L/A
r =  0(1+a(T-T0))
V=iR
E=J
Current and Resistance
Current and Resistance: Checkpoints,
Questions
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?