Lecture 22: WED 14 OCT Exam 2: Review Session CH 24–27.3 & HW 05–07 Physics2113 Jonathan Dowling Some links on exam stress:

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 , 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 q 2 /a for the work required to set up the four-charge configuration of Fig , assuming the charges are initially infinitely far apart. The electric potential at points in an xy plane is given by V = (2.0 V/m 2 )x 2 - (4.0 V/m 2 )y 2. 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 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: U 1 =0 Energy needed to bring in 2nd charge: Energy needed to bring in 3rd charge = Energy needed to bring in 4th charge = +Q Total potential energy is sum of all the individual terms shown on left hand side = So, final kinetic energy of each charge =

Potential V of Continuous Charge Distributions Straight Line Charge: dq=  dx  =Q/L Curved Line Charge: dq=  ds  =Q/2  R Surface Charge: dq=  dA  =Q/  R 2 dA=2  R’dR’ Straight Line Charge: dq=  dx  =Q/L Curved Line Charge: dq=  ds  =Q/2  R Surface Charge: dq=  dA  =Q/  R 2 dA=2  R’dR’

Potential V of Continuous Charge Distributions Straight Line Charge: dq=  dx  =Q/L Straight Line Charge: dq=  dx  =Q/L Curved Line Charge: dq=  ds  =Q/2  R Curved Line Charge: dq=  ds  =Q/2  R

Potential V of Continuous Charge Distributions Surface Charge: dq=  dA  =Q/  R 2 dA=2πR’dR‘ Straight Line Charge: dq=  dx  =bx is given to you. Straight Line Charge: dq=  dx  =bx is given to you.

✔ ✔ ✔ ✔

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+Q0 +V –V (a) True (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.

ΔxΔx (a) Since Δx is the same, only |ΔV| matters! |ΔV 1 | =200, |ΔV 2 | =220, |ΔV 3 | =200 |E 2 | > |E 3 | = |E 1 | The bigger the voltage drop the stronger the field. (b) = 3 (c) F = qE = ma accelerate leftward

Capacitors E =  /  0 = q/A  0 E = V d q = C V C plate =   0 A/d C plate =  0 A/d C sphere =  0 ab/(b-a) Connected to Battery: V=Constant Disconnected: Q=Constant Connected to Battery: V=Constant Disconnected: 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 (=  0 A/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 (=  0 A/d) Q=CV; Q decreases E = σ/  0 = Q/  0 A 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/C eq = ∑1/C j In parallel: same voltage C eq =∑C j

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 =  0 A/d Spherical: Cylindrical: C = 2  0 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 (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; V new = V (same) C new =  C (increases) Q new = (  C)V =  Q (increases). Since V new = V, E new = V/d=E (same) dielectric slab Energy stored? u=  0 E 2 /2 => u=  0 E 2 /2 =  E 2 /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; Q new = Q (same) C new =  C (increases) V new = Q/ C new = Q/ (  C) (decreases). Since V new < V, E new = V new /d = E/  (decreases) dielectric slab Energy stored? Example: Battery Disconnected — Voltage V Changes but Charge Q is Constant

Current and Resistance i = dq/dt V = i R E = J   =  0 (1+  (T  T 0 )) R =  L/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 3.5 x m. 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!

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

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 R eq =∑R j 1/C eq = ∑1/C j In parallel: same voltage same voltage 1/R eq = ∑1/R j C eq =∑C j

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

Problem: 27.P.018. [406649] Figure shows five 5.00 resistors. (Hint: For each pair of points, imagine that a battery is connected across the pair.) Fig (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+E -iR

(a) all tie (current is the same in series) The voltage drop is –iR proportional to R since i is same.

(a) V batt < 12V (walking with current voltage drop –ir (b) V batt > 12V (walking against current voltage increase +ir (c) V batt = 12V (no current and so ir=0)