physics 202, exam 1 review · physics 202, exam 1 review ... ∑=k q i i r i ∑ f=q e relation...
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Physics 202, Exam 1 Review
Logistics Topics: Electrostatics + Capacitors (Chapters 21-24)
Point charges: electric force, field, potential energy, and potential
Distributions: electric field, electric potential. Interaction of point charges with continuous distributions.
Conductors: charge distribution, electric field, electric potential
Capacitors: compute capacitance, energy stored in capacitor
Exam 1 Logistics
Exam time: Wednesday, February 17, 5:30-7PM
Rooms: 2103 Chamberlain and 2241 Chamberlain
2103: 304, 321, 310, 322, 305, 330, 324, 326, 307, 325, 301 2241: 302, 303, 327, 308, 329, 309, 323
Bring: Pen/pencil Calculator (no programming functionality)
1 single-sided formula sheet, self-prepared (no photocopying)
Exam 1: Electrostatics Topics, Mechanics
Topics: Coulomb Forces, Potential Energy Electric Field and Potential of Point Charges and Distributions Motion of charged particles in electric fields Electric field lines and equipotentials Conductors in electrostatic equilibrium Capacitance, Capacitors in circuits and dielectrics
Mechanics -- not the main focus, but you should know: Kinematics of uniformly accelerated particles Newton’s Laws: statics and dynamics Anything on homework is fair game for this midterm or the final
exam (e.g. circular orbits, springs)
Math -- you will not be expected to do nontrivial integrals. You should be able to do integrals which require simple substitutions.
Topics: Point Charges (I)
2 charges: force on q2 by q1
12r̂
principle of linear superposition
>2 charges: force on charge i
F21
F12
F21 F12
€
F i =
F 1i +
F 2i +
F 3i + ...
€
F 12 = k q1q2
r 2 r 12 = −
F 21
Topics: Point Charges (II)
Electric Field:
Electric Potential:
Field concept: electric field and electric potential
E =
Ei
i∑ = k qi
ri2 r̂i
i∑
V = Vii∑ = k qi
rii∑
F = q
E
Relation between force and field: VB −VA = −
E ⋅dl
A
B
∫
ΔU = qΔV
r = vector from source
to observation point
E = −
∇V
Gauss’s Law Net electric flux through any closed surface (“Gaussian
surface”) equals the total charge enclosed inside the closed surface divided by the permittivity of free space.
ΦE =
EidA∫ =
qenclε0
ε0: permittivity constant
Gaussian Surface
: all charges enclosed regardless of positions
electric flux qencl
Karl Friedrich Gauss
€
k =14πε0
Using Gauss’ Law Choose a closed (Gaussian) surface such that the surface integral is trivial. Use symmetry arguments:
1. Direction. Choose a Gaussian surface such that E is clearly either parallel or perpendicular to each piece of surface
2. Magnitude. Choose a surface such that E is known to have the same value at all points on the surface
EidA∫∫ = EdA∫∫ = E dA∫∫ = EA = qencl
ε0Then:
Given qencl, can solve for E (at surface), and vice versa
Continuous Charge Distributions (I)
Use Gauss’s Law to obtain E. Integrate to get V.
Examples: spherical symmetry, cylindrical symmetry, planar symmetry
Conductors (surface charge density only) and insulators
E ⋅dA∫∫ =
qenclε0
Method 1: high degree of symmetry
V (r) = −
E ⋅dr
ref
r
∫
Basic Symmetries
ΦE =
EidA∫∫ =
qenclε0
Use it to obtain E field for highly symmetric charge distributions.
Method: evaluate flux over carefully chosen “Gaussian surface”
spherical cylindrical planar
(point charge, uniform sphere, spherical shell,…)
(infinite uniform line of charge or cylinder…)
(infinite uniform sheet of charge,…)
Continuous Charge Distributions (II)
1. Direct calculation of E field: integrate to get V.
2. Direct calculation of V: take derivatives to get E.
Examples: uniformly charged ring, disk (on-axis), finite line charge.
dE = k dq
r2r̂
E = d
E∫
dV = k dqr V = dV∫
Conductors and Capacitors (I) Main feature of conductors:
Electrostatic equilibrium:
equipotentials
Application: Capacitors
E = 0 inside conductor
E = σε0
outside, perp to surface (only surface charges)
Capacitors Definition:
Computing capacitance: Parallel plate (also know: spherical, coaxial)
Capacitors in circuits: Charging/discharging Series and Parallel combinations
Q = CV
CP = Cii∑1
CS
=1Cii
∑
C =ε0Ad
€
U =12C
Q2 =12CV 2
Capacitors: Summary
• Definition:
• Capacitance depends on geometry:
d
A
- - - - - + + + +
Parallel Plates
a
b L
r
+Q
-Q
Cylindrical
a
b
+Q -Q
Spherical
C has units of “Farads” or F (1F = 1C/V) ε o has units of F/m
C ≡QΔV
C =εoAd
C = 4πεoabb − a
C =2πεoL
ln ba
⎛⎝⎜
⎞⎠⎟
Dielectrics • Empirical observation:
Inserting a non-conducting material (dielectric) between the plates of a capacitor changes the VALUE of the capacitance.
• Definition: The dielectric constant of a material is the ratio of the capacitance when filled with the dielectric to that without it:
κ values are always > 1 (e.g., glass = 5.6; water = 80) Dielectrics INCREASE the capacitance of a capacitor More energy can be stored on a capacitor at fixed voltage:
κ =CC0
ʹ′U =
CV 2
2=κC0V
2
2=κU
ε ≡ κε0permittivity:
€
E = E 0κ
κ
A capacitor of capacitance C holds a charge Q when the potential difference across the plates is V. If the charge Q on the plates is doubled to 2Q,
A. the capacitance becomes (1/2)V. B. the capacitance becomes 2C. C. the potential changes to (1/2)V. D. the potential changes to 2V. E. the potential does not change.
A capacitor of capacitance C holds a charge Q when the potential difference across the plates is V. If the charge Q on the plates is doubled to 2Q,
A. the capacitance becomes (1/2)V. B. the capacitance becomes 2C. C. the potential changes to (1/2)V. D. the potential changes to 2V. E. the potential does not change.
If C1 < C2 < C3 < C4 for the combination of capacitors shown, the equivalent
capacitance is A. less than C1. B. more than C4. C. between C1 and C4.
If C1 < C2 < C3 < C4 for the combination of capacitors shown, the equivalent
capacitance is A. less than C1. B. more than C4. C. between C1 and C4.
If C1 < C2 < C3 < C4 for the combination of capacitors shown, the equivalent
capacitance is A. less than C1. B. more than C4. C. between C1 and C4.
If C1 < C2 < C3 < C4 for the combination of capacitors shown, the equivalent
capacitance is A. less than C1. B. more than C4. C. between C1 and C4.
Two identical capacitors A and B are connected across a battery, as shown. If mica (κ = 5.4) is inserted in B,
A. both capacitors will retain the same charge. B. B will have the larger charge. C. A will have the larger charge. D. the potential difference across B will increase. E. the potential difference across A will increase.
Two identical capacitors A and B are connected across a battery, as shown. If mica (κ = 5.4) is inserted in B,
A. both capacitors will retain the same charge. B. B will have the larger charge. C. A will have the larger charge. D. the potential difference across B will increase. E. the potential difference across A will increase.
Example A capacitor is charged with a battery to a charge of Q=3.30×10-7 C. The area of the plates is 100 cm2=0.01 m2, and the distance between the plates is 1 cm=0.01 m. With dielectric inserted into the capacitor, the electric field is found to be 1.0×106 V/m.
1) What is the energy stored in the capacitor with the dielectric present?
Example A capacitor is charged with a battery to a charge of Q=3.30×10-7 C. The area of the plates is 100 cm2=0.01 m2, and the distance between the plates is 1 cm=0.01 m. With dielectric inserted into the capacitor, the electric field is found to be 1.0×106 V/m.
1) What is the energy stored in the capacitor with the dielectric present?
U = ½QV = ½Q(Ed) = ½(3.03×10-7)(3.03×10-7)(0.01) U = 1.52×10-3 J
Example A capacitor is charged with a battery to a charge of Q=3.30×10-7 C. The area of the plates is 100 cm2=0.01 m2, and the distance between the plates is 1 cm=0.01 m. With dielectric inserted into the capacitor, the electric field is found to be 1.0×106 V/m. The battery is disconnected and then the dielectric is removed, so the capacitor plates are separated by air. What is the energy stored in the capacitor after the dielectric has been removed?
Example A capacitor is charged with a battery to a charge of Q=3.30×10-7 C. The area of the plates is 100 cm2=0.01 m2, and the distance between the plates is 1 cm=0.01 m. With dielectric inserted into the capacitor, the electric field is found to be 1.0×106 V/m. The battery is disconnected and then the dielectric is removed, so the capacitor plates are separated by air. What is the energy stored in the capacitor after the dielectric has been removed?
Charge is same: Q=3.03×10-7 C Electric field without dielectric:
E0 =σε0
=Qε0A
=3.03×10−7
8.85 ×10−12( ) 0.01( )== 3.42 ×106 V / m
κ =EE0
=3.42 ×106
1.0 ×106= 3.42
U0 =κU = (3.42)(1.52 ×10−3) = 5.2 ×10−3J.