Exam 2: Review Session CH 24–27.3 & HW 05–07 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 (Ch 23) Gauss’s Law: Flux, Sphere, Plane, Cylinder (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.

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

Two Insulating Sheets

E does not pass through a conductor Two Conducting Sheets E does not pass through a conductor Formula for E different by Factor of 2 7.68 4.86 4.86 7.68 12.54

Electric Fields With Insulating Sphere

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 +Q +Q No energy needed to bring in first charge: U1=0 Energy needed to bring in 2nd charge: +Q +Q Energy needed to bring in 3rd charge = Total potential energy is sum of all the individual terms shown on left hand side = Energy needed to bring in 4th charge = So, final kinetic energy of each charge =

Potential V of Continuous Charge Distributions Straight Line Charge: dq=λdx λ=Q/L Curved Line Charge: dq=λds λ=Q/2πR Surface Charge: dq=σdA σ=Q/πR2 dA=2πR’dR’

Potential V of Continuous Charge Distributions Curved Line Charge: dq=λds λ=Q/2πR Straight Line Charge: dq=λdx λ=Q/L

Potential V of Continuous Charge Distributions Surface Charge: dq=σdA σ=Q/πR2 dA=2πR’dR‘ Straight Line Charge: dq=λdx λ=bx is given to you.

✔ ✔ ✔ ✔

ICPP: (a) True (b) False (c) True (d) True Consider a positive and a negative charge, freely moving in a uniform electric field. True or false? Positive charge moves to points with lower potential voltage. Negative charge moves to points with lower potential voltage. Positive charge moves to a lower potential energy. Negative charge moves to a lower potential energy. + + + + + + + + + – – – – – – – – (a) True –Q +Q +V –V (b) False (c) True (d) True

electron ✔ ✔ ✔ ✔

downhill

No vectors! Just add with sign. One over distance. Since all charges same and all distances same all potentials same.

(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

Capacitors E = σ/ε0 = q/Aε0 E = V d q = C V Cplate = ε0A/d Connected to Battery: V=Constant Disconnected: Q=Constant Csphere=ε0ab/(b-a)

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

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? 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. 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 SERIES: Q is same for all capacitors Total potential difference = sum of V 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 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 Disconnected — Voltage V Changes but Charge Q is Constant 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?

Current and Resistance i = dq/dt R = ρL/A Junction rule V = i R E = J ρ r = ρ0(1+a(T-T0))

Current and Resistance

A cylindrical resistor of radius 5. 0mm and length 2 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

Fill in the blanks. Think water in hose! 26.2: Electric Current, Conservation of Charge, and Direction of Current: Fill in the blanks. Think water in hose!

All quantities defined in terms of + charge movement! (a) right (b) right (d) right (c) right

DC Circuits Loop rule Multiloop Single loop V = iR P = iV

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

+E -iR (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) 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)