Lecture 22: WED 14 OCT Exam 2: Review Session

CH 24–27.3 & HW 05–07

Physics2113 Jonathan Dowling

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

2

2 1kQU QVL

= =•  Energy needed to bring

in 2nd charge:

2 2

3 1 2( )2

kQ kQU QV Q V VL L

= = + = +

•  Energy needed to bring in 3rd charge =

•  Energy needed to bring in 4th charge =

+Q +Q

+Q +Q Total potential energy is sum of all the individual terms shown on left hand side =

2 2

4 1 2 32( )

2kQ kQU QV Q V V VL L

= = + + = +

( )242

+LkQ

So, final kinetic energy of each charge = ( )24

4

2+

LkQ

Potential V of Continuous Charge Distributions

dV =kdqr

V = dV∫Straight Line Charge: dq=λdxλ=Q/LCurved Line Charge: dq=λdsλ=Q/2πRSurface Charge: dq=σdAσ=Q/πR2dA=2πR’dR’

r = ′R 2 + z2

Potential V of Continuous Charge Distributions

Straight Line Charge: dq=λdxλ=Q/L

Curved Line Charge: dq=λdsλ=Q/2πR

Potential V of Continuous Charge Distributions

Surface Charge: dq=σdAσ=Q/πR2dA=2πR’dR‘

Straight Line Charge: dq=λdxλ=bx is given to you.

(a) E-field does: + work?– work?

(c) Potential Energy: increase?decrease?

(b) Work done by me: + work?– work?

✔

✔

✔

+Vhigh = + + + + + + +

−Vlow = − − − − − − −

(d) Potential Voltage: higher?lower?

✔

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.

–Q +Q 0 +V

–V

(a) True (b) False (c) True (d) True

+ + + + + + + + +

– – – – – – – – E!"

(a) E-field does: + work?– work?

(c) Potential Energy: increase?decrease?

(b) Work done by me: + work?– work? ✔

✔

✔

(d) Potential Voltage: higher?lower?

✔

+Vhigh = + + + + + + +

−Vlow = − − − − − − −

−

electron

(−)

++++

−−−−

E!"

(a) → ? ← ? ↑ ? ↓ ?(b) 1= + 2 = + 3= + 4 = −

downhill

5 = +

(c) 3>1= 2 = 5 > 4

2

W1 = +10eVW2 = +10eVW3 = +20eVW4 = −10eVW5 = +10eV

D = 2d

(a) VP =+ed

+ +e2d

⎛⎝⎜

⎞⎠⎟ =

32ed= 3 / 2

For speed set d = e = 1!

(b) VP =+ed

+ +e2d

⎛⎝⎜

⎞⎠⎟ =

32ed= 3 / 2

(c) VP =+ed

+ +e2d

⎛⎝⎜

⎞⎠⎟ =

32ed= 3 / 2

VP(a) =VP

(b) =VP(c)

No vectors!Just add with sign.One over distance.

3

Since all charges same and all distances same all potentials same.

Va = +

Vc = −

Vb = 0

Va >Vc >Vb

Pa

Pc

Pb

4

Δx

Ex = dVdx

≅ ΔVΔx

(exact for constant field)

(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.

++++

++++

++++

−−−−

−−−−

−−−−

E!"

E!"

E!"

(b) = 3

−(c) F = qE = ma

accelerate leftward

ΔV ≡Vf −Vi

Capacitors E = σ/ε0 = q/Aε0 E = V d q = C V

Cplate = κ ε0A/d

Cplate = ε0A/d

Csphere=ε0ab/(b-a)

Connected to Battery: V=ConstantDisconnected: Q=Constant

Isolated Parallel Plate Capacitor: ICPP

•  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: (a) Increase? (b) Remain the same? (c) Decrease? If the plate separation is INCREASED, does the

Voltage V: (a) Increase? (b) Remain the same? (c) Decrease?

+Q –Q

•  Q is fixed! •  d increases! •  C decreases (= ε0A/d) •  V=Q/C; V increases.

dA

VQC 0ε==

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.

+Q –Q

•  V is fixed constant by battery! •  C decreases (=ε0A/d) •  Q=CV; Q decreases •  E = σ/ε0 = Q/ε0A decreases

Does the Electric Field Inside: (a) Increase? (b) Remain the Same? (c) Decrease?

dA

VQC 0ε==

AQE00 εε

σ ==

Battery does work on capacitor to maintain constant V!

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: C = ε0 A/d��

Spherical: • �Cylindrical: C = 2πε0 L/ln(b/a)]

Hooked to battery V is constant.

C = 114πε0

1a− 1b

⎡⎣⎢

⎤⎦⎥

Q=VC

(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

(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.

SERIES: •  Q is same for all capacitors •  Total potential difference = sum of V

•  Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E;

•  Battery remains connected•  V is FIXED; Vnew = V (same)•  Cnew = κC (increases)•  Qnew = (κC)V = κQ (increases).•  Since Vnew = V, Enew = V/d=E (same)

dielectric slab

Energy stored? u=ε0E2/2 => u=κε0E2/2 = εE2/2 increases

Example: Battery Connected — Voltage V is Constant but Charge Q Changes

•  Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E;

•  Battery remains disconnected•  Q is FIXED; Qnew = Q (same)•  Cnew = κC (increases)•  Vnew = Q/Cnew = Q/(κC) (decreases).•  Since Vnew < V, Enew = Vnew/d = E/κ

(decreases)

dielectric slab

Energy stored?

Example: Battery Disconnected — Voltage V Changes but Charge Q is Constant

u = ε0E2 / 2

→κε0Enew2 / 2 =κε0 E /κ( )2 / 2

= ε0E2 / κ 2( ) decreases

Current and Resistance i = dq/dt

V = i R E = J ρ

r = ρ0(1+a(T-T0))

R = ρL/A Junction rule

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 Ωm. 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

5A! "!

1+ 2 = 3

↑ 3A

3+ 2 = 5

↓ 2A2 + 4 = 6

6A! "!

8A! "!6 + 2 = 8

++++

−−−−

All quantities defined in terms of + charge movement!

i!(a) right

(c) right E!"→

(b) right J!"→

(d) right vd!"

→

DC Circuits

Single loop Multiloop

V = iR P = iV

Loop rule

Resistors vs Capacitors

Resistors Capacitors Key formula: V=iR Q=CV

In series: same current same charge Req=∑Rj 1/Ceq= ∑1/Cj

In parallel: same voltage same voltage 1/Req= ∑1/Rj 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

−→ +

(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 +)

+E -iR

i→

(a) all tie (current is the same in series)

(b) R1 > R2 > R3

The voltage drop is –iR proportional to R since i is same.

⇒ iR1 > iR2 > iR3⇒V1 >V2 >V3

(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)