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.