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Home , Electric charge

Lecture 11

Georg Simon Capp,,acitance, current, Ohm and resistance (1789-1854)

http://web.njit.edu/~sirenko/ 103 Spring 2012

2012 Andrei Sirenko, NJIT 1

Electric • Electric charge is fundamental characteristic of elementary particles •Two types of charges: positive/negative • Labels are simppyly a convention •Atomic structure : • negative cloud • nucleus of positive , uncharged

2012 Andrei Sirenko, NJIT 2 between pairs of point charges

+q1 F12 F21 −q2 or F12 +q1 +q2 F21 or F12 −q2 F21 −q1 ’s law -- the force between point charges: • Lies along the line connecting the charges. • Is proportilhional to the magni idfhhtude of each charge. • Is inversely proportional to the distance squared.

• Note that Newton ’ s third law says |F12|=|F| = |F21|!!

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Coulomb’s law

+q1 F12 F21 −q2

r12 For charges in a VACUUM 2 k | q || q | 9 N m 1 2 k = 8.99 ×10 2 | F12 |= 2 C r12

Often, we write k as: C 2 k = 1 with ε = 8.85×10−12 4πε 0 2 20120 Andrei Sirenko, NJITN m 4 Quantization of Charge • Charge is always found in INTEGER multiples of the charge on an electron/ • Electron charge e- = -1.6 10-19 [C] • Proton charge p = e = +1.6 x 10-19 Coulombs • Unit of charge: Coulomb (C) in SI units • One cannot ISOLATE FRACTIONAL CHARGE (e.g., 080.8 x 10-19 C1910C, +1.9 x 10-19 C, etc.) •e= Σ , (±2/3e, ±1/3e) • Charge: Q , q, -q, -575q, …., 7q, etc. • Q = 1 C is OK, it means Q = (1 ± 1.6 x 10-19) C

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Capacitance

• depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors +Q -Q • e.g. Area of conductors, separation, wh eth er th e space in between is filled (We first focus on with air,,p plastic , etc. where gap is filled by vacuum or air !) 2012 Andrei Sirenko, NJIT 6 Parallel Plate What is the electric in Area of each between the plates? (Gauss’ Law) plate = A σ Q Separation = d E = = ε ε A ch/harge/area = σ 0 0 = Q/A Relate E to potential difference V btbetween th e pl ltates: +Q -Q What is the capacitance C? ε A C = Q/V = = 0 ε =×8.85 10−12 F/ m d 0

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Isol ltdPated Parall lllPltel Plate C apacit or

• A parallel plate capacitor of capacitance C is charged using a battery. +Q -Q • Charge = Q, potential difference = V. • Battery is then disconnected. • If the plate separation is INCREASED, does potential difference V: (()a) Increase? (b) Remain the same? • Q is fixed!

(c) Decrease? • C decreases (=ε0A/d) • Q=CV; V increases. 2012 Andrei Sirenko, NJIT 8 Parallel Plate Capacitor & Battery • A parallel plate capacitor of cappggacitance C is charged using a battery. • Charge = Q, potential difference = V. • Plate separation is INCREASED while +Q -Q battery remains connected. • Does the inside: (a) Increase? (b) Remain the same? • V is fixed byyy battery! (c) Decrease? • C decreases (=ε0A/d) • Q=CV; Q decreases

• E = Q/ ε0A decreases

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Example 2 What is the potential difference across each capacitor?

• QQ;Q = CV; Q is same for all ca pacitors 10 µF 20 µF 30 µF • Combined C is given by: 1 1 1 1 = + + Ceq (10µF) (20µF) (30µF) 120V

•Ceq = 5.46 µF • Q = CV = (5.46 µF)(120V) = 655 µC Note: 120V is shared in the •V1= Q/C1 = (655 µC))(/(10 µF) = 65.5 V ratio of INVERSE •V2= Q/C2 = (655 µC)/(20 µF) = 32.75 V •V3= Q/C3 = (655 µC)/(30 µF) = 21.8 V i.e.1:(1/2):(1/3) (largest C gets small est V)

2012 Andrei Sirenko, NJIT 10 Example 3 Same potential

Which of the following A statements is FALSE? (a) B, D and F are in PARALLEL. B (b) E and F are in SERIES. C D (c) A, C and E are in PARALLEL.

E F

120 V 2012 Andrei Sirenko, NJIT 11

Example 4 A If each capacitor = 10 µF, what is the charge on C D B capacitor A?

• A, C, E are in parallel • B, D, F are in parallel E F

120 V • Po tenti al diff erence across each set of capacitors = 120V/2 = 60V • Potential difference across A = 60V • Charge on A = (10 µF)(60 V) = 600 µC 2012 Andrei Sirenko, NJIT 12 Example 5 In the circuit shown, what 10 µF is the charge on the 10µF capacitor?

5 µF 5 µF 10V • The t wo 5µF capacitors are i n parallel • Replace by 10µF 10 µF • Then, we have two 10µF capacitors in series • So, there is 5V across the 10µF 10 µF capacitor of interest 10V • Hence, Q = (10µF )(5V) = 50µC

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Energy Stored in a Cap acitor

• Start out with uncharged capacitor • Transfer small amount of charge dq from one plate to the other until charge on each plate has dq magnitude Q • How much was needed?

Q Q q Q2 CV 2 U = ∫Vdq = ∫ dq = = 0 0 C 2C 2 2012 Andrei Sirenko, NJIT 14 Example 6

• 10µF capacitor is initially charged to 120V . 10µF(CF (C1) 20µF capacitor is initially uncharged. • Switch is closed, equilibrium is reached. • How much is dissipated in the process? Initial charge on 10µF = (10µF)(120V)= 1200µC 20µF (C2) After switch is closed, let charges = Q1 and Q2. Charge is conserved: Q + Q = 1200µC 1 2 • Q = 400µC Q Q Q 1 Also, Vfinal is same: 1 = 2 2 •Q = 800µC Q1 = 2 C1 C2 2 •Vfinal= Q1/C1 = 40 V 2 2 Initial energy stored = (1/2)C1Vinitial = (0.5)(10µF)(120) = 72mJ 2 2 Final energy stored = (1/2)(C1 + C2)Vfinal = (0.5)(30µF)(40) = 24mJ Energy lost (dissipated) = 48mJ Question: where did the 2012 Andrei Sirenko, NJITenergy go?? 15

Summary for Capacitors

1 1 1 • Capacitors in SERIES: = + +... (same Q) Ceq C1 C2

•Capacitors in PARALLEL: Ceq = C1 + C2+… (same V) • Capacitors store energy: U = (1/2)CV2=Q2/2C

2012 Andrei Sirenko, NJIT 16 Georg Simon Ohm (1789-1854)

"a professor who preaches such heresies is unworthy to teach science.” Prussian minister of education 1830

CtditCurrent and resistance

Microscopic view of charge flow: , resistivity and drift speed Macroscopic view: current and resistance Relating microscopic and macroscopic Conductors, semiconductors, insulators, superconductors

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Conductors in electrostatic equilibrium Suppose a piece of copper wire is placed in a static electric field and not connected to anything else. What happens? • Naïve view: are free to move; they arrange themselves in such a way that E = 0 inside the wire; no more Random motion of electrons: electron motion! • Movement is random because of “collisions” with • Reality: electrons inside the wire keep vibrating nuclei moving all the time, but on average they • In between collisions, arrange thlhemselves so th hE0iid!at E = 0 inside! electrons move VERY FAST ~106 m/s!

2012 Andrei Sirenko, NJIT 18 Conductors in absence of equilibrium

Non-equilibrium -- imagine that you attach a battery across a copper wire so that electrons can be put into the wire and extracted from the wire: • Now: E is NOT ZERO inside thdhe conductor • Electrons “drift” because of the non-zero electric field (“”) E • Drift is much, much • Drift of electrons creates a net SLOWER than random flow of charge -- CURRENT motion: typically < mm/s measured in Amperes = C/s dQ • Convention: “current” is viewed I = [A = C / s] as flow of POSITIVE charge dt 2012 Andrei Sirenko, NJIT 19

Current: example 1

• The figure shows charges moving at the given rates +5 C/s -5C/s5 C/s • What i s th e total current flowing through the area -1 C/s shown? • Remember that current is the flow of POSITIVE charge! • TotalAAl current = 5A + 5A - 1A +5 C/s = 9A (towards the right) +5 C/s +1 C/s

2012 Andrei Sirenko, NJIT 20 Current: Example 2 • Charge is CONSERVED: total I current into junction = total 2 current out of junction I1

• In figure shown, e.g. I1=I2+I3 • Useful analogy: think of water flow! I3 • Example: what is the current I ihin the wi re sh own at ri ih?ght? 3A3 A 1 A

• Answer: I = 3A 1A1 A 2A2 A I = ?

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Resistance • Since electrons collide with vibratingg, nuclei, there is “resistance” to the free flow of Current = I charge. • Define “Resistance”= R = V/I (i.e. “how much charge flow do I get for an applied potential difference?”) Potential difference = V • Units: Ohm (Ω) V • Macroscopic property -- e.g. we R = talk ab out “ resi stance” of an I object such as a specific piece of wire. This is not a local property Circuit symbol for R 2012 Andrei Sirenko, NJIT 22 Relating the Macro- and Microscopic views • Conductor of Current = I UNIFORM cross- sectional area A, length L Potential difference = V •Potentildiffial difference V V EL ρL applied across ends R = = = • No te: wh a t if th e area I JA A was not uniform?? (HW problem)

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Resistance:

How m uch pot enti al do I n eed to app ly to a dev ice to d rive a g ive n current through it? L E L R = ρ = A J A Ohm’s laws V V Georg Simon Ohm R ≡ and therefore : i = and V = iR (1789-1854) i R

Volt Units : [R] = ≡ Ohm (abbr. Ω) For many materials, R remains a constant for a wide range of values of current and potential. Devices specifically designed to have a constant value of R are called resistors, and symbolized by

2012 Andrei Sirenko, NJIT 24 Variation of resistance (resistivity) with temperature

R, ρ change with temperature in a complicated, material-dependent way.

Why d oes it ch ange? N ucl ei vib rat e due to thermal agitation, and scatter electrons as they pass.

For many conductors, it can be approximated by a linear temperature dependence (for a small range of temperatures),

ρ − ρ0 = α(T − T0 ) ρ0 With α determined empirically and listed in tables.

Trivia: why do light bulbs mostly die at the moment of switch-on? Answer: when the filament is cold it has less resistance, therefore it is the momentht when th e currentit is maxi mum.

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Various materials and temperature:

ρ Conductors: Examples: copper , aluminum, iron. 0 T ρ Semiconductors

Examples: 0 Germanium, T Heike Kammerlingh- silicon. Onnes ρ 0.1 to 20K for “usual” materilials, e.g. mercury (1911) Karl Mueller Superconductors Johnannes Bednorz 30 to 180K for “high T : 0 C temperature” materials, T e.g. YBaCu (1986) Tc 2012 Andrei Sirenko, NJIT 26 Power dissipation: Resistance was a measure of the “cost” of establishing a current in a realistic conductor. The “cost” can be characterized in terms of the energy one needs to constantly input to a conductor in order to keep a current going.

+ Let us follow an amount of charge dq as it moves b c through the circuit, starting at a. V R From a to b, through the battery, its is increased by Vdq. a d From b to c its potential is constant, similarly from c to d.

When it is back at a, its potential energy should be the same as when it started. Therefore there must have been a loss of potential energy of amount -Vdq when moving through the resistance. dU dU = Vdq = V idt ⇒ Power = = V i dt Units: Watt 2 2 ⎛V ⎞ V Applying Ohm's laws : Power = (i R)i = i R Power = V ⎜ ⎟ = 2012 Andrei Sirenko, NJIT⎝ R ⎠ 27 R

Summary:

• We saw that chargggges moving through conductors experience “resistance” to their motion. • resistance is related to the electron drift speed • We discussed microscopic and macroscopic view of electrical currents. • Studied temperature and material dependence. • Discussed how moving charges costs and delivers electrical power.

2012 Andrei Sirenko, NJIT 28 Lecture QZ

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Summary:

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