Physics 235 Exam 1

a=F/m=qE/m=9.6x10^13m/s²: ANSWER: B The electron is accelerating much more than the proton, so use the electron mass. a=F/me=ke²/mer²=7.02*10^22m/s². Answer A.

1. A proton is moving in a uniform electric field of 10⁶V/m. What is its acceleration? A. 9.8m/s² B. 9.6*10^13m/s² C. 1.8*10^9m/s² D. 1.7*10^3m/s² E. 1.0*10^-3m/s² 2. What is the centripetal acceleration of an electron in circular orbit about a proton with an orbital radius of 0.60Å? (Constants are on the front of the test.) A. 7.0×10^22m/s² B. 9.8m/s² C. 4.6×10^4m/s² D. 1.7×10^17m/s² E. 2.4×10^19m/s²

1. The work is QV=1.2mC*1.2V=1.44mJ. Do not divide by time, which would give you POWER (watts). ANSWER: D. 2. The energy lost is QV=7.0C*1.5V=10.5J . Answer B. 3. The energy provided is QV=24 J (or power times time, i.e. IVt, but not IV.) Answer A.

1. A resistor is connected across a 1.2V battery for 10 seconds. In those 10 seconds 1.2mC flow through the resistor. How much energy is dissipated in the resistor? A. 14.4mJ B. 10mJ C. 7.2mJ D. 1.44mJ E. 0.72mJ 2. A resistor is connected across a 1.5V battery for 10 seconds. In those 10 seconds 7.0C flow through the resistor. How much energy is dissipated in the resistor? A. 14.4J B. 10.5J C. 7.2J D. 1.44J E. 0.72J 3. During a 10 sec interval, a charge of 2.0C passes through a device hooked up to a 12V battery. How much electrical energy does the battery provide? A. 24J B. 12J C. 120J D. 0J E. 2J

1. R₃₄ = 2.5Ω. This is in series with R₁ and R₂. The total is R₁₂₃₄ = 12.5Ω. So I₁₂₃₄ = 10V/12.5Ωs = 0.8A. This is also the current through R₁, since it's a series combination. The power consumed in 1 is: P₁=I1²R₁|=(0.8A)²(5Ω)=3.2W. Answer A. 2. R₁ and R₂ are in series. R₃ and R₄ are in parallel. First combine R₃ and R₄ to get the equivalent R₃₄=2Ω. Now R₁, R₂, and R₃₄ are in series with a total resistance of 4Ω+4Ω+2Ω=10Ω. The current through the battery is 10V/10Ω=1A and the power is P=IV=10W. Ans: E

1. Diagram question 2. In the circuit below the resistances are R₁=R₂=R₃=R₄=4Ω. What is the power provided by the battery? A. 3.2W B. 1.6W C. 8.0W D. 4.0W E. 10W

1. The capacitance doesn't change. The voltage difference, surface charge density, and electric field are all proportional to the magnitude of the charge and all double. Answer E. 2. Does not affect the capacitance! The electric field, surface charge density and voltage difference are halved. Answer B.

1. If the charge of a parallel plate capacitor is doubled: A. the electric field is halved B. the capacitance is halved C. the capacitance is doubled D. the surface change density is not changed on either plate E. the electric field is doubled 2. If the charge of a parallel plate capacitor is halved: A. the capacitance is doubled B. the electric field is halved C. thecapacitanceishalved D. the surface change density is not changed on either plate E. the electric field is doubled

1. The electric field points to lower potential, i.e. inside. Its magnitude is E=∆V/∆X=1.2*10⁷V/m. Answer D. 2. The electric field is E=−ΔV/ΔL (Volts per meter) to lower potential, the exterior in this case. The magnitude is 5.0MV/m. Answer C.

1. In its normal resting state the potential difference across the axon membrane is 85mV; the potential inside is lower than outside. What is the electric field within the membrane if the thickness is 7nm? A. 8.5×10³V/m toward the exterior of the cell B. 1.2×10⁷V/m toward the exterior of the cell C. 8.5×10⁹V/m toward the interior of the cell D. 1.2×10⁷V/m toward the interior of the cell E. Zero because the cell membrane is a conductor. 2. During an action potential, the electric potential of the inside of the axon rises briefly to +40mV above the exterior. What is the electric field within the membrane at that moment if its thickness is 8nm? A. 3.2×10³V/m toward the interior of the cell B. 3.2×10³V/m toward the exterior of the cell C. 5.0×10⁶V/m toward the exterior of the cell D. 3.2×10⁹V/m toward the interior of the cell E. 5.0×10⁶V/m toward the interior of the cell

1. The equivalent resistance of the 4 circled resistors is: (1/Req)=(1/25Ω)+(1/15Ω)+(1/10Ω+15Ω)⇒ Req=6.81Ω. That equivalent resistor is in series with the 45Ω resistor. The current delivered by the battery (and through the 45Ω resistor) is I45Ω=V45Ω/45Ω=45V/45Ω=1.00A That current will also flow through the "eq" circled combination. Veq=IReq=6.81V. Finally, this voltage is the voltage across the 25Ω resistor, so: I25Ω/25Ω=6.81V/25Ω=0.27A. Answer A. 2. See the current and voltage in the diagram above. The three top branches are in parallel and their voltage difference is 0.20*25=5V. The current in the middle branch is Imid=5V/15Ω=0.33A. The current in the bottom branch is Ibottom=5V/(10Ω+15Ω)=0.20A. At the junction the three incoming current combine to flow to the left I45Ω=I35Ω=Ibatt=0.20+0.33+0.20=0.733A. The voltage across the 45Ω resistor is V45=I45(45Ω)=33V. Answer B.

1. In the circuit shown, the voltmeter, (V) reads 45V. What does the ammeter (A) read? A. 0.27A B. 35A C. 0.007A D. 1A E. 15A 2. In the circuit shown, the ammeter (A) reads 0.20A. What does the voltmeter (V) read? A. 55V B. 33V C. 45V D. 72V E. 9.0V

1. Since we know everything in the upper branch, I'd take a path from b to a (remember ∆V=Vf-Vi) along that route. This will travel with the 1A current. ∆V=-(1A)(6Ω)+20V-(1A)(1Ω)=+13V. Answer B. 2. The current junctions required that 1A go (to the left) through the middle row, so 1A goes through ε₁1. The power is I₁ε₁=18 W. ANSWER D 3. (-1A)(6Ω)+20V-(1A)(1Ω)-(2A)(1Ω)-ε₂-(2A)(2Ω)=0 --> -6V+20V-1V-2V-ε₂-4V=0 ε₂=20-6-1-2-4=7J Answer D

1. In the circuit shown, what is potential at a minus the potential at b, i.e. Va−Vb ? A. -7V B. +13V C. +33V D. +72V E. +9.0V 2. In the circuit shown, ε₁=18 V and ε₂= 7 V. What is the power provided by ε₁? A. 25W B. 36W C. 14W D. 18W E. 7W 3. In the circuit shown, what is EMF for the battery in the lowest branch i.e., ε₂? A. 9.0V B. 3.5V C. 5.0V D. 7.0V E. 1.0V

1. The initial kinetic energy of the proton becomes all potential at closest approach: Ki=Uf=kq1q2/r=k(e)(42e)/(6*10^−15)=10 MeV. Answer D. 2. We want KEi=-Uf=+ke²/r=15.12*10⁶eV=15.12 MeV. Ans: B 3. U=ke²/r=9.6*10⁵eV, where I converted to eV by leaving out one factor of e. Answer D. 4. Same |F|-->Newton's 3rd law. a=F/m Answer B.

1. The nuclear reaction 100Mo+1H→99Tc+2H requires that protons (1H) are accelerated to overcome the electric repulsion from the 100Mo nucleus. The charge of the proton is qp=e and Qmo=42e. What is the minimum proton kinetic energy required to allow the proton to come within 6*10^-15 m and allow the nuclei to react? (The Mo nucleus remains essentially stationary) A. 6.0eV B. 2.72eV C. 8×10-19eV D. 10MeV E. 6.0keV 2. To produce 99Tc an isotope used in medical imaging studies, protons are accelerated to an energy sufficient to bring them within r=4x10^-15m of a 100Mo nucleus. The charge of the proton is +e and the charge of the 100Mo nucleus is 42e. What is the minimum initial kinetic energy of the proton required for it to come to within the distance r of the 100Mo nucleus? (Recall that KEi+Ui=KEf+Uf and that Ui=0.) A. 4.5eV B. 15.2MeV C. 1.5x10²eV D. 42meV (4.2x10-2eV) E. 13.6eV 3. How much work is needed to assemble an atomic nucleus containing two protons (Helium)? The distance between the protons in the nucleus is 1.5 × 10^−15 m. Assume the protons started from very far away. A. 1.9 MeV B. 6.0 eV C. 8×10-19 eV D. 0.96 MeV E. 6.0 keV 4. In the planetary model of the Hydrogen atom the electron orbits the proton. Why is it reasonable to treat the proton as stationary while the electron orbits around it? A. The motion of the electron and the proton is all relative. The force of the acceleration on the proton keeps it centered. B. The proton and the electron feel the same magnitude force but the acceleration of the electron is far greater because of its lower mass. C. The proton and the electron feel different forces proportional to their mass and thus the electron is the one that accelerates the most. D. The electron feels a much smaller magnitude force than the proton. But because the electron has a much smaller mass it moves more than the proton. E. The proton feels a much smaller force than the electron and so it does not move appreciably compared to the electron.

1. U=qV=e*(-84 mV)=-84mEV (Ans: B) 2. U=qV=+eV=-84meV. Answer B.

1. The resting potential difference from the outside to the inside of an axon is ΔV=−84mV (the potential is lower inside the axon). What is the potential energy of a K+ ion located inside the axon if zero potential energy is defined as zero outside the axon? A. U= -134meV B. U= - 84meV C. U= +134meV D. U= +84meV E. U= -13.6meV 2. The resting potential inside an axon is −84mV. What is the potential energy of a Na+ ion located inside the axon? A. -134meV B. - 84meV C. +134meV D. +84meV E. -13.6meV

1. Read off the resistance at 120K, R≈7.2mΩ. P=I²R=1.8mW. Answer B. 2. R≈10mΩ. V=IR=(0.5A)(10mΩ)=5mV. Answer C.

1. What is the approximate power dissipated in the material at 120K (-153°C)? A. 120W B. 2mW C. 8mW D. 0W E. 0.5mW 2. What voltage would be measured across the resistor at 240K? A. 40Ω/K B. 10V C. 5mV D. 92K E. 0V

1. Do the loop! Answer B. 2. You could use the loop rule: F = (∆PR+∆PL)/(Rs+Rp) The heart sides work together against both resistances. Solve F to get answer C. 3. Loop rule (starting from lower right corner and going with the current): +VL−IRs+VR−IRp=0 → VR=IRs+IRp−VL=20V. Answer A.

1. Which equation represents the loop rule for our model of the circulatory system? A. ∆VR+∆VL+FRs+FRp=0 B. ∆VR+∆VL-FRs-FRp=0 C. ∆VR-∆VL+FRs-FRp=0 D. ∆VR+∆VL-FRs+FRp=0 E. ∆VR-F∆VL-FRs-FRp=0 2. Which equation gives the flow rate for our model of the circulatory system? A. (∆PR+∆PL)/(Rs-Rp) B. (∆PR-∆PL)/(Rs-Rp) C. (∆VR+∆VL)/(Rs+Rp) D. (∆PR+∆PL)*(Rs-Rp) E. (∆PR-∆PL)*(Rs+Rp) 3. Diagram question

This is an MP question. The capacitance afterward is C=kC0=52uF. The energy is U=1/2CV². Plug in numbers to get 17.6mJ. Answer C.

A capacitor has (originally) a capacitance of 13.5μF. It is connected to a power supply that keeps a constant potential difference of 26.0V across the plates. A piece of material having a dielectric constant κ=3.85 is then placed between the plates completely filling the space between them. How much energy is stored in the capacitor after the dielectric is inserted? A. 145J B. 9.2mJ C. 17.6mJ D. 5.35J E. 89mJ

Charge is fixed, so U is proportional to 1/C. Further, C is proportional to k. Since the dielectric is removed, the capacitance decreases, so the energy increases. Answer A.

A dielectric is inserted between the plates of a capacitor. The system is then charged by a battery. The battery is disconnected and the dielectric is removed. How does the electrostatic energy stored in the capacitor change when the dielectric is removed? A. Increases B. Decreases (but not to zero) C. Stays the same D. It depends on the value of the plate separation E. Decreases to zero

The minimum energy (in direction) is when the dipole is aligned with the field. The field is due to +Q. At the location of p that field is directed in the -y direction. Answer D.

A positive charge +Q is located on the y-axis, as shown. An electric dipole, p, is located at the origin. (The dot represents the position of p, not its orientation.) What orientation of p will have the minimum energy? Recall that p=ql, where l is the displacement vector from the negative to the positive charge of the dipole. A. p directed along the +y axis B. p directed into the page C. p directed along the -x axis D. p directed along the -y axis E. p directed out of the page

As the second charge moves, it get close to -Q. The E-field magnitude increases, V becomes more negative, i.e. decreases, and the potential energy increases (because the moving charge is negative too.) Answer D.

A small negatively charged sphere (point charge -Q) is located at a fixed position as shown. A second NEGATIVE charge moves from point A to point B. Which of the following indicates how the magnitude of the electric field, the electric potential (V) and the electric potential energy of the negative charge (U) change in moving from A to B?

B is further from the negative charge so magnitude of Field decreases and V becomes more positive (V increases). U=qV with q positive, so U increases as well. Ans: A

A small negatively charged sphere (point charge -Q) is located at a fixed position as shown. A second POSITIVE test charge moves from point A to point B. Which of the following indicates how the magnitude of the electric field, the electric potential (V) and the electric potential energy of the negative charge (U) change in moving from A to B?

The +Q charge creates the field and potential. Here the test charge is moving farther from the source of the field and potential. The field magnitude |kQ/r²| is going down. The potential, kQ/r, is decreasing as r increases. The change in potential energy is ΔU = qΔV is increasing, since the test charge, q, is negative. Answer B

A small positively charged sphere (point charge +Q) is located at a fixed position as shown. A second NEGATIVE test charge moves from point A to point B. Which of the following indicates how the magnitude of the electric field, the electric potential (V) and the electric potential energy of the negative charge (U) change in moving from A to B?

The negative charge moves closer to the positive charge. The electric field and potential are created by Q. So the field and V increase because it's closer to a positive charge. The potential energy decreases because the second charge is negative and thus more closely bound. Answer B.

A small positively charged sphere (point charge Q) is located at a fixed position as shown. A second NEGATIVE charge is moved from point A to point B. Which of the following indicates how the magnitude of the electric field, the electric potential (V) and the electric potential energy of the negative charge (U) change in moving from A to B?

U=Uab+Ubc+Uac=(ke(-2e))/d+(ke(-2e))/d+ke²/2d =ke²/d(-4+1/2)=-3.5ke²/d =-45.8 V Answer B.

A. -15.8 eV B. -45.8 eV C. -35.8 eV D. -55.8 eV E. -25.8 eV

You can do this vector sum to get the answer when D>>d. But for a diploe the strength of the field is proportional to p/r³=qd/D³ and since F=QE the force on the charge Q must be proportional to qQd/D³ too. The dipole creates a field to the right at the location of Q so the force is to the right too. Answer D. Using Couloumb's law we add the two forces, one from each charge in the dipole. The two forces are equal in magnitude, kqQ/D², and their vertical components cancel. The right components add. The right component is 2kqQsin(θ)/D², (the two coming from the sum of both contributions). This is (2kqQ/D²)Sin(θ) = (2kqQ/D²)(d/2)=kqQd/D³. Answer D.

An electric dipole is made of a positive charge q and a negative charge -q separated by a small distance d. This dipole is placed near a positive point charge Q in the position shown in the diagram, so that each of the dipole charges is a distance D from the point charge. What is the electric force on the charge Q?

The voltage at the location of the charge Q (due to the dipole) is zero, so the potential energy is zero. In mathematical language: Uq=QV=Q((kq/D)+k(-q)/D))=0 Answer A.

An electric dipole is made of a positive charge q and a negative charge -q separated by a small distance d. This dipole is placed near a positive point charge Q in the position shown in the diagram, so that each of the dipole charges is a distance D from the point charge. What is the electric potential energy of the charge Q (do not include the binding energy of the dipole)?

The force on -Q is F = -QE, where E is the field from the dipole at the location of -Q. The field is to the left, so the force is to the right. The magnitude of the field from a dipole is proportional to p/r³ = ql/r³. Answer D. In a more complete fashion, the field a distance r from a dipole along the dipole axis is 2kp/r³, so the force is (in the dipole approximation) 2kpQ/r³, which is answer D.

An electric dipole is made of a positive charge q and a negative charge -q separated by a small distance l, oriented as shown in the picture. A negative point charge -Q is a distance r from the center of the dipole, such that all three charges lie on a line. The distance r >>l. Which is most nearly the magnitude and direction of the electric force on the charge -Q? Hint: You might identify the correct answer, without actually computing it, by eliminating incorrect answers.

Veel=0.1v. When in series, total V=Veel*#eel Answer E.

Eels generate potential differences of 400V to stun enemies and prey. These potentials are produced by cells that each can generate 0.1V. How should the cells be connected together to produce 400V? A. 4000 cells connect in parallel B. 40 cells connected in series C. 200 cells connected in parallel D. 40 cells connected in parallel E. 4000 cells connected in series

The negative charge experience a leftward force and the positive an equal rightward force. The net force is zero and the torque is CCW. Answer E. (Or the dipole moment will rotate to align with E.)

If an electric dipole is in a region of constant electric field, as shown, what will happen initially to the dipole? A. It will fly off to the right B. It will fly off to the left C. It will remain fixed in place D. It will rotate clockwise in this picture. E. It will rotate counter-clockwise in this picture

I=V/R=0.92A. Answer D.

If the body has a resistance of 130Ω and a voltage of 120V (normal household voltage) is applied across the body, what is the current that would flow through the body? A. 120A B. 60kA C. 4A D. 920mA E. 500A

1. The tube resistance is proportional to 1/A² so reducing A by a factor of 2 INCREAESES R by a factor of 4. Using Ohm's law: I=V/R, the increased R for constant pressure (V) REDUCES the flow (I) by a factor of 4. Ans: E. 2. Pressure differences are like voltage difference. Volume flow is like charge flow and resistance is ... resistance. ∆P=FRtube. Answer D. 3. Answer C.

In coronary artery disease, the area of the arteries decreases due to the build up of plaque. If the area of a particular artery decreases by a factor of two, how does the blood flow change if the total pressure drop across the artery remains constant? (Recall Rtube=8πηL/A².) A. It stays the same B. It increases by a factor of 2 C. It decreases by a factor of 2 D. It increases by a factor of 4 E. It decreases by a factor of 4 Which of the following equations represents the application of Ohm's Law (∆V=IR) to blood flow? P is pressure, η is viscosity, F is volume flow rate, Rtube is fluid resistance, and W is work. A. ∆F = PRtube B. ∆W = PRtube C. ∆P=Fη D. ∆P = FRtube E. ∆F=WRtube Diagram question.

From Newton's 3rd law, the force of electron on nucleus is equal in magnitude to the force of nucleus on electron. Answer: C

In the Li+ ion, one electron (charge -e) is circling the nucleus, which has a charge +3e. Which of the following statements about the electric force is true? A. The magnitude of the force on the electron is 3 times the force on the nucleus. B. The magnitude of the force on the electron is 1/3 times the force on the nucleus. C. The magnitude of the force on the electron is equal to the force on the nucleus. D. The system has positive electric potential energy (U(infinity)=0). E. The system is unbound.

Loop rule: +q/C-iR=0 --> I=q/RC. Answer A.

In the circuit below, with the switch closed and the direction of the current indicated, what is the current through the resistor for a charge q on the capacitor? A. I=q/RC B. I=-q/RC C. I=qRC D. I=-q/C E. I=-qC/R

Do the loop: 16V-I*(1.4Ω +4Ω+1.6Ω)−8V-I*1.6Ω=0 8V-I*16Ω=0 I=0.5A (Ans: A)

In the circuit shown, the two realistic batteries have internals resistances as indicated, and positive current I flows clockwise as shown. What is I? A. 0.5A B. 1A C. 3A D. 4A E. 10A

Start at the (closed) switch and go around counter-clockwise: From - to + terminal (+ε); with the current (-iR); from + to - plate (-Q/C); back to the start (0). Answer A.

In the circuit shown, what is the correct loop equation with the switch closed?

84mV=10^-7A*2*10⁶Ω(1-e^(-t/RC))=200mV*(1-e^(-t/RC)) 0.42=(1-e^(-t/RC)) e^(-t/RC)=0.58 -t/RC=ln(0.58) --> t=-ln(0.58)RC=1.1µs Answer A.

In the circuit we presented to represent charge flow in nerve conduction, a stimulus "closes the switch" and a constant current I0 flows and charges the capacitor. As we showed, the charge on the capacitor is: q(t)=I₀RC(1-e^(-t/RC)). For C=1pF and R=2MΩ and I₀=100nA, about how long will it take for the capacitor to charge to 84 mV? (remember V=q/C) A. 1.1μs B. 1s C. 32s D. 2ns E. 84ms

I=ΔQ/Δt=6.3x10⁷*1.6x10^-19 C/10^-3s=10nA. (Ans: B)

Nerve signals propagate due to an influx of Na+ ions, each with charge +e, from outside into an axon. Typically about 6.3×10⁷ Na+ ions enter each millimeter of the axon in about 1ms. What is the approximate electric current for 1mm of axon over this period? A. 63μA B. 10nA C. 1A D. 22mA E. 400μA

Add up the potential energies of the three pairs. U=(Uleft-mid)+(Uleft-right)+(Umid-right) =(k(-e)(2e)/d)+(k(-e)(-e)/2d)+(k(-e)(2e)/d)=-7ke²/2d Plugging in numbers gives: −2.7×10^−17J=-168eV. Answer E.

The Helium atom, a bound system, has two electrons, each with charge -e, and a nucleus of charge +2e. The average distance between the electrons is 0.6Å (6×10−11m) and the average distance between the electrons and nucleus is d = 0.3Å, so this can be modeled effectively by the picture shown. What is the electrical potential energy of the three-body helium atom? A. +168eV B. ZERO C. +13.6eV D. -13.6eV E. -168eV (recall 1eV = 1.6*10^-19J)

The resistance is R axon=ρL/A=ρL/πr²=3.2MΩ. The capacitance is C=(C/A)2πrL=1.57pF (around the cylinder). The time constant is approximately the RC time: τ=RC=5.0=10^−6s.

The capacitance per unit area and resistivity ρ of a mammalian axon are given below. Use our capacitor model of an axon as a cylinder of radius r=10μm (10^-5m) and length 0.5mm (5×10^−4m) to estimate the time constant τ for depolarizing the axon membrane. The resistivity is ρ=2.0-mΩ; Capacitance per unit area: C/A=5*10^−5F/m². A. 0.5s B. 7ms C. 5μs D. 2s E. 3×10⁷s

The energy stored is 1/2QV=1/2Q²/C (V=Q/C). U increases by 4x when Q increases by 2x. Ans: D

The capacitor in a primitive (DC) defibrillator provides the charge for a "jolt" of current that can reset the heart rhythm. For a current of 20A in 10ms at 500V, the energy stored in the capacitor is U=50J. What energy would be stored for twice as much charge (40A in 10ms)? A. Half as much (25J) B. The same (50J) C. Twice as much (100J) D. Four times as much (200J) E. One tenth as much (5J)

The capacitor will charge until the current I=0 (capacitor charge does not change, i.e. equilibrium). Then the loop equation is: V-IR-Q/C=V-Q/C=0 (for I=0) or Q=CV=10^-12F*0.1V=10^-13 Ans: A.

The circuit shown models a portion of an axon with very large cell-wall resistance. The voltage provided by the battery is V=100mV, the resistance is R=10MΩ and the capacitance is C=1pF. What is the maximum charge on the capacitor? A. 10^-13C B. 10^-8C C. 10^-11C D. 10^-15C E. 10^-5C

V=kQ/r OR C=Q/V=r/k r=4*10⁷m/2π=6.4*10⁶ V=1.4kV Answer A

The earth is a reasonably good conductor. The circumference of the Earth is about 40,000km. If the net charge on the earth is 1C, what is the electric potential of the Earth (with the potential defined to be zero very far from the Earth)? (It might be helpful to recall how we estimated the capacitance of a person.) A. 1.4kV B. 140V C. 14V D. 1.4V E. 14kV

The former is exponential decay. The latter is exponential growth. Answer D.

The equation dQ/dt=-Q/t represents a capacitor discharging in a circuit. What could be represented by this equation: dQ/dt=+Q/t for finite (non-zero) τ? A. Radioactive decay such as 14C after an organism dies B. Equilibrium where the birth and death rates are the same C. Controlled growth with the death rate greater than the birth rate D. Exponential growth such as an explosion E. Growth to a steady state as in a leaky bucket

1. An electron moving towards a negative charge is HIGHER POTENTIAL ENERGY. The change in potential is -150V. For an electron with charge -e, the change is +150eV. (No need to multiply by e and then divide by e.) ANSWER: B 2. The field points toward the negative charge, i.e. right. The magnitude is |∆V/∆s|=30V/10cm=300V/m. Answer D.

The figure shows three charges with equipotential lines in 50V increments and a grid with 1cm x 1cm boxes, so points A and B are 2.5cm apart. 1. What is the change in potential energy of an electron moving from point A to point B? A. -150eV (the electron has LOWER potential energy) B. 150eV (the electron has HIGHER potential energy) C. -6keV (the electron has LOWER potential energy) D. 6keV (the electron has HIGHER potential energy) E. 0 (the electron potential energy doesn't change) 2. What is the approximate electric field midway between points A and B? A. 0V/m B. 30V/m to the right C. 30V/m to the left D. 300V/m to the right E. 300V/m to the left

Since U=qV and q=-e, the highest potential energy occurs where the potential is lowest, point A. Alternately, the electron is very close to another negative charge at point A. The electric field is perpendicular to equipotential lines and points to lower potential. Answer up and to the left.

The image shows three charges, two positive and one negative, as indicated in the figure. The lines represent lines of equipotential. 1. At which of the above labeled points would an electron have the highest potential energy? 2. What direction would an arrow point if it represented the electric field vector at point E?

To the left of +Q the field is to the left. To the right of +4Q the field is to the right. Between the charges the +Q contribution is right and the +4Q contribution is left. Pick a point a distance x to the right of +Q. The field is 0=+kQ/x²-k(4Q)/(d-x)² or kQ/x²=k(4Q)/(d-x)². Cross multiply, take the square root, divide by kQ and get d−x=2x. Answer d/3. Check if it makes sense: close to the +Q than the +4Q. Answer B.

Two charges +4Q and +Q are placed a distance d apart as shown. Where is the electric field zero? A. d/2 to the right of +Q B. d/3 to the right of +Q C. d to the left of +Q D. 2d to the left of +Q E. 3d to the left of +Q

The electric field contribution (at A) due to the -Q charge is kQ/x² to the right and the contributions from the +4Q charge is (k(4Q)/(d+x)²) to the left, so for the net field to be zero: (k(4Q)/(d+x)²=kQ/x²⇒4x²=(d+x)²⇒x=d. Answer E.

Two charges +4Q and -Q are placed a distance d apart as shown. The electric field is zero at point A. FIND x. A. x=2d B. x=d/2 C. x=d/3 D. x=3d E. x=d

In region A the contribution to E from +Q points left. The contribution from −2Q points right. This region is closer to the smaller magnitude charge, so the contributions from the two can be balanced. Answer A. In region B, C, D the field is to the right. In region E the field is to the left because the contribution from −2 will exceed the contribution from +Q.

Two charges +Q and −2Q are separated as shown. Indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is zero. (C is at the midpoint between the charges.)

U=k*(e*-e)/(2x10^-9 m)+ k(e*e)/(2.5*10^-9 m) = -0.72eV+0.58eV=-0.14eV (Ans: C) (Pythagorean theorem gives 2.5nm)

What is the potential energy of the protein charge-dipole bond in the orientation shown below. The dipole p consists of charges +e and -e separated by 1.5nm (p=2.4x10^-28C-m), and the charge is +e. The distance from the charge to the dipole's negative charge is 2nm. Do not include the potential energy stored in the dipole. A. U=-72eV B. U=480MeV C. U=-0.14eV D. U=1J E. U=0

Count the energy pair-wise for all three pairs. U=(kq(-2Q)/r)+(kQ(Q)/2r)+(k(-2Q)Q/r) = -4+1/2)kQ²/r=-7Q²/2r. Answer D.

What is the total electrical potential energy of the arrangement of charges shown below? You may take the potential energy when the three charges are infinitely far apart to be zero.

No current anywhere. Answer C.

What would happen in this circuit if the light bulb A burned out, creating a break in the wire at that location? A. Bulb A would turn off and bulbs B and C would dim, but remain lit B. Bulb A would turn off and bulbs B and C would remain at the same brightness C. Bulb A would turn off and bulbs B and C would turn off too D. Bulb A would turn off, bulb B would brighten, and bulb C would dim E. Bulb A would turn off and bulbs B and C would brighten

Answer D.

Which of the following most accurately explains the physical phenomena represented in this equation? A. Sodium ions are positively charged and chlorine ions are negative B. The capacitance of the axon is proportional to the area of the axon membrane C. Electrical resistivity is a function of temperature D. The combination of thermally driven diffusion and ion electric potential energy E. The balance of electric and magnetic forces