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These are the notes of Past Paper of Physics for Scientist and Engineers and its key important points are: Isolated System, Conversion Factors, Permittivity of Vacuum, Charge of Electron, Universal Gas Constant, Specific Heat of Water, Gauss Law
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Answer all six problems. Write clearly and explain your work. Partial credit will be given for incomplete solutions provided your logic is reasonable and clear. Cross out any parts that
be credited. If you obtain an answer that is questionable, explain why you think it is wrong.
Constants and Conversion factors Some useful equations
Avogadro number, NA 6. 022 × 1023 Permittivity of vacuum, 0 8. 85 × 10 −^12 F·m−^1 Permeability of vacuum, μ 0 4 π × 10 −^7 T·m·A−^1 Charge of electron, qe = −e − 1. 602 × 10 −^19 C Mass of electron, me 9. 11 × 10 −^31 kg Universal gas constant, R 8.315 J·mol−^1 ·K−^1 = 1.99 cal·mol−^1 ·K−^1 Boltzmann constant, k 1. 381 × 10 −^23 J·K−^1 Stefan-Boltzmann constant, σ 5. 67 × 10 −^8 W·m−^2 ·K−^4 Acceleration due to gravity, g 9.8 m·s−^2 Specific heat of water 1 kcal·kg−^1 ·◦C−^1 Heat of fusion of water 80 kcal·kg−^1 1 atm 1. 013 × 105 N·m−^2 1 kcal 4. 18 × 103 J 1 hp 746 W
Coulomb′s law : F =
4 π 0
q 1 q 2 r^2
ˆr
Electric f ield : dE =
4 π 0
dq r^2
ˆr
Electric dipole : p = qd T orque on a dipole : ~τ = p × E P otential energy of a dipole : U = −p · E
Gauss′s law :
E · dA =
qencl 0
P otential dif f erence : Vab = −
∫ (^) b
a
E · dl
P otential : dV =
4 π 0
dq r P otential energy : Uab = qVab Electric f ield and potential : E = −∇V Capacitance : C =
q Vab
Table 1: Dielectric Constants for Some Materials Material Dielectric Constant Dielectric Break Down K (V/m)
Vacuum 1.0000 > 1015 Air 1.0006 3 × 106 Teflon 2.1 60 × 106 Paraffin 2.2 10 × 106 Polystyrene 2.6 24 × 106 Vinyl (plastic) 2 - 4 50 × 106 Paper 3.7 15 × 106 Quartz 4.3 8 × 106 Oil 4 12 × 106 Glass Pyrex 5 14 × 106 Porcelain 6 - 8 5 × 106 Mica 7 150 × 106 Silicon Dioxide 3.9 5 − 15 × 106 Silicon 11.68 − × 106 Water (liquid) 80 — Strontium titanate 300 8 × 106
Table 2: Resistivity for Some Materials Material Resistivity Temperature Coefficient ρ (Ω · m) α(◦C)−^1 at 20◦C Conductors Copper 1. 59 × 10 −^8 0. Silver 1. 68 × 10 −^8 0. Gold 2. 44 × 10 −^8 0. Aluminum 2. 65 × 10 −^8 0. Tungsten 5. 6 × 10 −^8 0. Iron 9. 71 × 10 −^8 0. Platinum 10. 6 × 10 −^8 0. Mercury 98 × 10 −^8 0. Nichrome 100 × 10 −^8 0. Semiconductors Carbon (graphite) (3 − 60) × 10 −^5 -0. Germanium (1 − 500) × 10 −^3 -0. Silicon 0.1- 60 -0. Insulators Glass 109 − 1012 Hard Rubber 1013 − 1015 Silicon Dioxide 1014 − 1016
(a) [10 points] Circle T or F for True or False
T F (i) The total electric charge of an isolated system remains constant regardless of changes within the system itself.
T F (ii) A plasma is a poor conductor.
T F (iii) Any two opposite charges make a dipole.
T F (iv) The superposition principle for electric fields is well confirmed by experiment.
T F (v) A insulator’s electrons are not free to move under electrical force.
T F (vi) Gauss’s law is always valid.
T F (vii) The inverse of resistance and resistivity is conductance and conductivity.
T F (viii) The electric field is a vector field with SI units of Newtons per Coulomb (N/C) or, equivalently, volts per meter (V/m). The strength of the field at a given point is defined as the force that would be exerted on a positive test charge of +1 Coulomb placed at that point; the direction of the field is given by the direction of that force.
T F (ix) The capacitance of a collection of charges is determined solely by the geometry of the charge distribution.
T F (x) Electric charge is a characteristic of subatomic particles, and is quantized when expressed as a multiple of the so-called elementary charge e = 1. 6 × 10 −^19 C.
T F (xi) An electric battery is composed of two dissimilar metals (or appropriate material such as carbon) and a solution called an electrolyte.
T F (xii) In order to have a steady-state electric current one needs a complete circuit.
T F (xiii) Conductivity plus Resistivity of a material is equal to one.
T F (xiv) In the microscopic model of electric current the validity of Ohm’s law requires the electrons moving with a high velocity and drifting with a much lower one.
T F (xv) The lethal electric shock is around 0.1 Amperes (100 mA) for a period of a second or more.
T F (xvi) The electric flux is how much electric field passes through a surface per unit time.
T F (xvii) The electric potential is equal to the electric potential energy.
T F (xviii) A capacitor can store electric charge.
T F (xix) Electrical Resistivity depends upon Temperature.
T F (xx) Kirchhoff’s Rules are really only conservation of charge and energy.
(A) Field lines cannot cross each other. (B) The field lines should turn sharply as you move from one charge to the other. (C) The field lines should be smooth curves in vacuum. (D) The field lines should always end on negative charges or at infinity
(v) In the figure below, the electric field lines are shown for a system of two point charges, QA and QB. Which of the following could represent the magnitudes and signs of QA and QB? In the following, take q to be a positive quantity.
(A) QA = + q and QB = −q. (B) QA = +7q and QB = − 3 q. (C) QA = +3q and QB = − 7 q. (D) QA = − 3 q and QB = +7q. (E) QA = − 7 q and QB = +3q. (F) All are correct.
(vi) Several electric field line patterns are shown in the diagrams below. Which of these patterns are incorrect? Circle the letter of the incorrect patterns and explain what is wrong with all incorrect diagrams.
(vii) There are several ways to produce electrostatic charging. Which is not one? (A) Rubbing two dissimilar materials together. (B) Induction. (C) Contact. (D) Radio broadcast. (E) Electric Sparking. (F) None cause electrostatic charging.
(viii) A neutral conducting sphere of radius r is located a distance d from a dust particle of charge q. The conducting sphere exerts a force on the charged particle which is (A) zero. (no net force) (B) repulsive. (C) attractive. (D) not enough information to determine.
(ix) A fixed potential difference V exists between a pair of close parallel plates carrying opposite charges +Q and -Q. Which of the following would not increase the magnitude of charge that you could put on the plates? (A) Increase the size of the plates. (B) Move the plates farther apart. (C) Fill the space between the plates with paper. (D) Increased the fixed potential difference V. (E) None of the above.
(x) Sharper Image and other producers makes substantial profit from selling ion generators (electrode creates negative ions in the air) as air cleaner and antibacterial agent. How does generating negative ions clear the air of dust and bacteria? (A) The ions are attracted to the dust and bacteria by polarizing them and make them charged. The dust is then attracted to any polarizable surface and sticks. (B) The ions are attracted to the dust and bacteria and chemically react to disassemble them. (C) The ions are attracted to the dust and bacteria and make them effectively charged. Since the Earth is positively charged by the large thunderstorms in the Amazon, the dust and bacteria are attracted to the floor. (D) The coronel discharge that makes the negative ions is like a spark and actually burns up the dust and bacteria. (E) It does not really work but the negative ions smell like fresh bleach so that people think the air is cleaner.
(c.3) [2 points] What is an expression for the potential V as a function of radius r? Set V (r = 0) = 0.
(c.4) [2 points] What is the stored energy per unit length in the system at voltage V?
(c.5) [2 points] If the shaded region in between the cylinders is filled with a dielectric material with constant K while keeping the charge per unit length fixed, how do the quantities calculated in parts c.2, c.3, and c.4 change? Briefly describe your reasoning.
(d) [10 points] Two spheres placed far apart are positively charged to the same potential (as they are connected by a thin conducting wire) as seen in Figure 2. The radius rs of the smaller spherical end is about half of the radius rl of the larger end. (d.1) [4 points] Sketch in the figure the distribution of the positive charges and the electric field lines. (See question sections below first.)
Figure 2: Figure for problem 1(d) Two conducting spheres one with twice the diameter as the other and the two are connected by a conductor.
Explain your reasoning: (d.2) [2 points] Give an expression for the charge qi on each sphere for potential V.
(ii) What is the electric field at the radius, r = 0. 53 × 10 −^10 m, from the center of a proton? (A) 5 × 106 rˆ N/C (B) 5 × 107 rˆ N/C (C) 5 × 108 rˆ N/C (D) 5 × 109 ˆr N/C (E) 5 × 1010 rˆ N/C (F) 5 × 1011 ˆr N/C (G) − 5 × 106 ˆr N/C (H) − 5 × 107 rˆ N/C (J) − 5 × 108 ˆr N/C (K) − 5 × 109 rˆ N/C (L) − 5 × 1010 rˆ N/C (M) − 5 × 1011 rˆ N/C
(iii) What is the electric potential at the radius, r = 0. 53 × 10 −^10 m, from the center of a proton, if V (∞) = 0? (A) 6.8 V (B) 13.6 V (C) 27.2 V (D) 54.4 V (E) -6.8 V (F) -13.6 V (G) -27.2 V (H) -54.4 V
(iv) What is the electric field strength near a large flat insulating disk with surface charge density σ = 10−^4 C/m^2? (A) 5. 6 × 106 N/C (B) 5. 6 × 107 N/C (C) 5. 6 × 108 N/C (D) 5. 6 × 109 N/C (E) 5. 6 × 1010 N/C (F) 5. 6 × 1011 N/C
(iv) The dipole moment of a water molecule is 6. 1 × 10 −^30 C · m. If the distance between the two hydrogen atoms and the oxygen atom is 10−^10 m in the direction of the resultant dipole, what is the net positive charge on each hydrogen atom? (A) 1e = 1. 6 × 10 −^19 C (B) 0.8 e (C) 0.6 e (D) 0.4 e (E) 0.2 e
(v) The dipole moment of a water molecule is 6. 1 × 10 −^30 C ·m. What is the torque on a water molecule molecule if its dipole vector is at right angles to an electric field of E = 5 × 104 N/C (A) 3 × 10 −^30 N · m (B) 3 × 10 −^25 N · m (C) 3 × 10 −^20 N · m (D) 0 (E) 1. 2 × 10 −^36 N · m
(vi) What is the electric flux through a sphere encircling a free electron? (A) 1. 81 × 10 −^10 N m^2 /C (B) 1. 81 × 10 −^9 N m^2 /C (C) 1. 81 × 10 −^8 N m^2 /C (D) 1. 81 × 10 −^7 N m^2 /C (E) 1. 81 × 10 −^6 N m^2 /C (F) − 1. 81 × 10 −^10 N m^2 /C (G) − 1. 81 × 10 −^9 N m^2 /C (H) − 1. 81 × 10 −^8 N m^2 /C (I) − 1. 81 × 10 −^7 N m^2 /C (J) − 1. 81 × 10 −^6 N m^2 /C
(vii) A Gaussian surface with one portion inside of a conductor encloses a flux of 2 × 105 N m^2 /C. How much surface charge on the conductor is enclosed by the Gaussian surface? (A) 1. 77 × 10 −^6 C (B) 1. 77 × 10 −^5 C (C) 1. 77 × 10 −^4 C (D) 1. 77 × 10 −^3 C (E) 1. 77 × 10 −^2 C (F) 1. 77 × 10 −^1 C (G) 1. 77 C (H) 17. 7 C
(b) [3 points] Circle correct answer The Earth’s Field (i) The Earth has an electric field which has an average magnitude of 150 N/C near its surface. The field points radially inward. The radius of the Earth is 6371 km. What is the net charge of the Earth? (A) 7C (B) 68C (C) 677C (D) 6771C (E) 6. 8 × 104 C (F) 6. 8 × 105 C (G)
(ii) The Earth has an electric field which has an average magnitude of 150 N/C near its surface. The field points radially inward. The radius of the Earth is 6371 km. What is the voltage of the Earth? (Assume zero potential at infinity.) (A) 10^9 V (B) 10^6 V (C) 10^3 V (D) 0V (E) − 103 V (F) − 106 V (G) − 109 V
(iii) A human standing on the surface of the Earth is at the same potential as the surface of the Earth. According to the figure, what is the best description of the effective electrical behaviour of a human? (A) an insulator (B) a good conductor (C) a poor conductor (D) a semiconductor (E) a superconductor (F) a lightening rod
Figure 3: Figure for problem 2(b)(iii) showing human standing on the Earth.
(d) [5 points] Circle correct answer (i) The lowest AC current that can cause ventricular fribillation is? (A) 1 mA (B) 10 mA (C) 100 mA (D) 1 A (E) 10 A
(ii) If the internal resistance of a person is 200 Ohms over a distance of 0.2 m and area of 0.1 m by 0.1 m, then what is the resistivity ρ inside a human? (A) 1 Ω · m (B) 10 Ω · m (C) 100 Ω · m (D) 1000 Ω · m (E) 4000 Ω · m
(iii) What is the electric field in a medium with resistivity ρ = 1 Ω · m carrying a current density of J = 10A/m^2? (A) 0.1 V/m (B) 1 V/m (C) 10 V/m (D) 100 V/m (E) 1000 V/m (F) 0.01 V/m
(iv) If the current density is J = 10A/m^2 moves at 60◦^ to an area of A = 0. 01 m^2 , what is the total current through A? (A) 1 mA (B) 5 mA (C) 10 mA (D) 50 mA (E) 100 mA
(v) If the peak of AC current is 15 A, what is the RMS current Irms? (A) 1 A (B) 10.6 A (C) 15 A (D) 21.2 A (E) 120 VAC
(e) [5 points] Circle correct answer (i) A biological battery called an electroplaque produces an EMF of 0.15 V and has an internal resistance of 0.25 Ohms. What net voltage does this electroplaque produce for a 20 mA current? (A) 0.01 V (B) 0.05 V (C) 0.10 V (D) 0.145 V (E) 0.2 V
(ii) If four 10-Ohm resistors are put in parallel, what is the equivalent resistance? (A) 0.25 Ohm (B) 1 Ohm (C) 2.5 Ohms (D) 10 Ohms (E) 40 Ohms
(iii) If four 10 pF capacitors are put in series, what is the equivalent capacitor? (A) 2.5 pF (B) 10 pF (C) 25 pF (D) 40 pF (E) 100 pF (F) 250 pF
(iv) If four 10 Volt batteries are put in parallel, what is the equivalent battery voltage? (A) 2.5 V (B) 4 V (C) 10 V (D) 25 V (E) 40 V
(v) If the charge on a fixed capacitor is doubled, the energy stored in the capacitor (A) halves. (B) is unchanged. (C) doubles. (D) quadruples. (E) triples. (F) is quartered.
(b) [7 points] As of 1980, the longest cost-effective distance for electricity transmission was 4,000 miles (7,000 km), although all present transmission lines are considerably shorter. It operated as a +/-300,000 V DC power line. The voltage is boosted to keep it constant. The resistance for the electrical transmission wire is about 0.31 Ohms/km. (b.1) Find a formula for the power loss Ploss as a function of the power transmitted, P the total voltage difference V and the resistance R = 0. 31 L, where L is the distance.
(b.2) If the transmission losses were 20% for this longest distance transmission, what was the delived power?
(b.3) Highest Voltage Systems The current highest transmission voltage (DC) is +/- kV on HVDC Itaipu (Brazil). China is building a very high voltage power transmission system that will transmit electricity at a DC voltage of +/- 800 kV. At the same time this project will have a power transmission capacity of 5000 MW = 5 Gigawatts, by running 4 wires in parallel for each voltage. The system is scheduled to commence commercial service by mid-2010. In future the electricity generated by several hydro-electric power plants will be transported from Yunnan via 1,400 km to Guangzhou over this long-distance HVDC link. What is the current and how much power is lost in the transmission? If the cross-sectional area of the wire were halved relative to that of part (b) because copper is so much more expensive now, then what would be the losses?
(c) [8 points] Superconducting cables High-temperature superconductors promise to revolutionize power distribution by providing lossless transmission of electrical power. The development of superconductors with transition temperatures higher than the boiling point of liquid nitrogen has made the concept of superconducting power lines commercially feasible, at least for high-load applications. It has been estimated that the waste would be halved using this method, since the necessary refrigeration equipment would consume about half the power saved by the elimination of the majority of resistive losses. Such cables are particularly suited to high load density areas such as the business district of large cities, where purchase of a wayleave for cables would be very costly. In 2006 Chubu University in Japan performed an experiment with a superconducting cable that could carry a current of 2,200 A and was insulated to withstand 20 KV operating at a temperature in range of 72 K to 80 K. (c.1) How much power could this superconducting cable carry?
(c.2) What is the power cost of refrigeration per unit length assuming that at 80 K the re- frigeration consumed half the power lost in a comparable (0.31 Ohm/km) normal conductor?
(c.3)If the cable can carry 20% more current, when cooled to 40 K, what is the maximum power and the net gain in power transmitted (i.e. account for the extra refrigerator power)? Assume that the heat leak is unchanged going to the lower temperature and the refrigerator efficiency keeps the same ratio to ideal and TH = 300K.
