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Unformatted text preview: PHY138Y – Electricity and Magnetism – Quarterly Test Solutions
February 8, 2005 Multiple Choice Questions
1. Two point charges, Q and q, are a distance r apart. The electric ﬁeld is zero at a point P between the charges, on the line segment connecting them. Then (A) Q and q must have the same magnitude and sign; (B) Q and q must have the same sign, but may have different magnitudes; (C) Q and q must have equal magnitudes and opposite signs; (D) Q and q must have opposite signs, and may have different magnitudes; (E) the point P must be midway between Q and q. Remember the electric ﬁeld points away from a positive charge and towards a negative charge. If the charges have opposite sign (choices C and D here) then some electric ﬁeld lines will point directly from the positive charge to the negative charge in the region in between. If Q and q have the same sign and magnitude then there will be a zero electric ﬁeld half way between the charges (choices A and E) so these are possible solutions – but because there are two of them you should have realised you need to go deeper. In fact the need not have the same magnitude, nor does the point where = 0 be half way in between. As long as the charges have the same sign then there will be a point with = 0 somewhere E E along the line segment connecting the charges ⇒ B is the right answer. 2. An electron is placed in a uniform electric ﬁeld directed toward the right side of the page. What is the direction of the force that ﬁeld exerts on the electron? (A) toward the right; (B) toward the left; (C) upwards; (D) downwards; (E) perpendicular to the plane of the page. Remember the electric ﬁeld direction is deﬁned as the direction a positive charge would be pushed. Thus the electron, that has a negative charge, will be pushed in the opposite direction: bf to the left. 3. The number of electric ﬁeld lines per unit area is: (A) directly proportional to the ﬁeld strength; (B) inversely proportional to the ﬁeld strength; (C) directly proportional to the square of the ﬁeld strength; (D) directly proportional to the cube of the ﬁeld strength; (E) inversely proportional to the square of the ﬁeld strength. This is just the deﬁnition of how we learned electric ﬁeld lines were drawn. It is the number of lines per unit area, in three dimensions, despite the fact that we usually only draw these in two dimensions. The third dimension is always understood. 4. A solid dielectric ﬁlls the space between the plates of a parallelplate capacitor. This capacitor and a resistor are connected in series across the terminals of a battery. Now the plates of the capacitor are pulled slightly farther apart leaving a small air gap (as shown in the ﬁgure). When equilibrium is restored in the circuit, (A) the potential difference across the plates has increased; (B) the energy stored on the capacitor has increased; (C) the capacitance of the capacitor has increased; (D) the charge on the plates of the capacitor has decreased. The capacitor is connected (through a resistor, but that doesn’t matter because the question suggests we wait until “equilibrium is restored”) to a battery. So the potential between the plates will not change – A is not the answer. When the capacitor plates are pulled apart the capacitance decreases (C = εA/d ) – C is not the right answer. The fact that there is a dielectric between the plates would complicate the actual calculation of the capacitance value, but it will still go down. It takes a little more effort to narrow this down further. U = CV 2 /2, but V is constant (see above) so if the capacitance goes down the U will go down – B is not the answer. Finally Q = CV , so if V remains constant and C goes down the Q will go down – D is the answer. 5. The equivalent capacitance in pF for the network shown when C = 100 pF is: (A) 58; (B) 0.38; (C) 68; (D) 0.46; (E) 150. The 3C and rightmost C capacitors are in parallel, so they can be combined to give a equivalent capacitor of 4C. Then one is left with 2C, 4C, and C in series: Cequiv = [1/2C + 1/4C + 1/C]−1 = 4/7C With C = 100 pF, this gives Cequiv = 57.1 pF – the unconventional rounding notwithstanding, the best answer is A. 6. Two wire segments made of the same material have the same length but different diameter. They are connected in series to a battery. The quantity that is the same for the two wires is: (A) the potential difference across the length of each segment; (B) the current in each segment; (C) the current density in each segment; (D) the electric ﬁeld in each segment. This was very similar to one of the ’quick quizzes’ given in class. The key to ﬁnding the answer here is to remember that because the wire segments are made of the same material they will have the same resistivity, but since they have different diameters, the one with the large diameter will have a smaller resistance. The battery will generate a current through the combined wires (connected in series) that is the same in each segment. Since the resistances are different there will be different voltage drops (A is not the right answer) and different electrics ﬁelds (D is not the right answer). Since the current density is related to the electric ﬁeld by the resistivity and the resistivity is the same, but the electric ﬁeld is not the current densities will not be the same either (C is not the right answer). In fact – as mentioned above – the currents will be the same and so B is the right answer. 7. Given a 6 Ω resistor, a 9 Ω resistor and an 18 V battery, what is the maximum power (in W) that can be dissipated? (A) 104; (B) 71; (C) 90; (D) 80; (E) 22. The resistors can be connected to the battery in two ways: in series or in parallel. If they are connected in series there will be an equivalent resistance of 15 Ω and P = V 2 /R = 182 /15 = 21.6 Watts will be dissipated in the circuit. If the resistors are connected in parallel then there will be an equivalent resistance of Requiv = [1/6 + 1/9]− 1 = 3.6Ω. This leads to a power of P = 182 /3.6 = 90 W. This is the largest power that can be dissipated in this circuit so C is the right answer. 8. Three identical resistors are connected across a voltage source V so that one of them is in parallel with two others which are connected in series. The power dissipated through the ﬁrst one, compared to the power dissipated by each of the other two, is approximately (A) the same; (B) half as much; (C) twice as much; (D) four times as much. The effective circuit here will have Re f f = R connected to the voltage V in parallel with Re f f = 2R (two resistors identical to the ﬁrst in series). They will have the same voltage drop across them so we can use P = V 2 /Re f f to compute the power dissipated in each branch. Since the resistance is twice as large in the branch with two resistors the total power dissipated there will be half as much as in the ﬁrst brach/ﬁrst resistor. BUT the question asks how much power is dissipated in “each of the ... two”. Despite several requests for clariﬁcation, I think the English as unambiguous here. So there will be onequarter as much power dissipated in each of the paired resistors – compared to the lone resistor in the other branch – or the ﬁrst resistor will dissipate four times as much power as either of the twin resistors in the second branch of the circuit. D is the right answer. 9. A 100 Ω resistor and a 10 µF capacitor are connected to a 300 V battery. At time t = 0, the battery is switched out of the circuit, leaving only the capacitor and resistor. 4 ms after the switch closes, the current in the resistor (in mA) is: (A) 6; (B) 18; (C) 406; (D) 55; (E) 2336. This was a relatively straightforward RC circuit question. The time constant in this circuit is 100Ω · 10µF which gives 1 ms. So after 4 ms the current will have decayed from its initial value (I0 = V /R = 300/100 = 3 A) by a factor of e−4 = 0.0183. So the current will be 54.9, or 55 mA. The answer is D. Worked Problem:
10. The outer surface of a cell membrane (of dielectric constant κ ≡ εε0 = 6.0 and thickness 10 nm), is charged positively, through an excess of Sodium (Na+ ) ions, and the inner portion is charged negatively, through an excess of Chlorine (Cl − ) ions. You may assume that the excess charges on the inner and outer surfaces of the cell membrane are equal in magnitude. The voltage drop across the membrane is ΔV = 90 mV. Making the reasonable assumption that the cell membrane can be treated as a parallel plate capacitor of plate area 4.0 × 10−9 m2 : NB. All answers should have had two signiﬁcant ﬁgures, despite some queries to the contrary, 10 has two signiﬁcant ﬁgures (1 ×101 would be the way to write this with one signiﬁcant ﬁgure). But we accepted three signiﬁcant ﬁgures. (1) (10 points) Determine the capacitance of the membrane. C = εA/d = 6.0 × 8.85pF/m · 4.0 × 10−9 m2 /10 × 10−9 m = 21.2pF C = 21 pF (2) (7 points) What is the charge on the outer surface of the membrane? Q = CV = 21.2pF · 90mV = 1.91pC Q = 1.9 pC (3) (6 points) How many excess Sodium (Na+ ) ions are there on the outer surface ? The charge on each Na+ ion is 1.6 × 10−19 C so there are: NNa+ = 1.91pC/(1.6 × 10−19 C) = 11.9 × 106 ions on the surface of the cell. NNa+ = 12 × 106 ions (4) (7 points) Determine the magnitude of the electric ﬁeld inside the membrane. E = V /d = 90mV/10nm = 90/10 × 106 = 9.0MV/m   = 9.0 MV/m E (5) (7 points) How much energy is stored by the cell in the electric ﬁeld established in the membrane? U = CV 2 /2 = 1 21.2pF(90mV)2 / = 0.0859pJ 2 Electrical Energy = 0.086 pJ ...
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This note was uploaded on 01/20/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Charge, Magnetism

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