MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics: Final Exam Review Session Problems Solutions

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1 Department of Physics: 8 Problem 1: Spherical Capacitor 8 Final Exam Review Session Problems Solutions A capacitor consists of two concentric spherical shells The outer radius of the inner shell is a = 1 m and the inner radius of the outer shell is b = m a) What is the capacitance C of this capacitor? b) Suppose the maximum possible electric field at the outer surface of the inner shell before the air starts to ionize is E max (a) = V m -1 What is the maximum possible charge on the inner capacitor? c) What is the maximum amount of energy stored in this capacitor? Also show that Q / C = u E dv d) When E(a) = V m -1 what is the potential difference between the shells? Solution: The shells have spherical symmetry so we need to use spherical Gaussian surfaces Space is divided into three regions (I) outside r b, (II) in between a < r < b and (III) inside r a In each region the electric field is purely radial (that is E = Eˆr ) Region I: Outside r b: Region III: Inside r a : These Gaussian surfaces contain a total charge of, so the electric fields in these regions must be as well Region II: In between a < r < b : Choose a Gaussian sphere of radius r The electric flux on the surface is E da = EA = E 4πr

2 Department of Physics: 8 The enclosed charge is Q enc = +Q, and the electric field is everywhere perpendicular to the surface Thus Gauss s Law becomes E 4πr = Q ε E = Q 4πε r That is, the electric field is exactly the same as that for a point charge Summarizing: Q E = 4πε r ˆr for a < r < b elsewhere We know the positively charged inner sheet is at a higher potential so we shall calculate ΔV = V (a) V (b) = a b E d a Q s = 4πε r dr Q = 4πε b r b a = Q 1 4πε a 1 b > which is positive as we expect We can now calculate the capacitance using the definition C = Q ΔV = Q Q 1 4πε a 1 = 4πε 1 b a 1 b = 4πε ab b a C = 4πε ab b a = (1 m)( m) (9 1 9 N m C )(1 m) = 1 11 F Note that the units of capacitance are ε times an area ab divided by a length b a, exactly the same units as the formula for a parallel-plate capacitor C = ε A / d Also note

3 Department of Physics: 8 that if the radii b and a are very close together, the spherical capacitor begins to look very much like two parallel plates separated by a distance d = b a and area A 4π a + b 4π a + a = 4πa 4πab So, in this limit, the spherical formula is the same at the plate one 4πε C = lim ab b a b a ε 4πa d = ε A d b) Suppose the maximum possible electric field at the outer surface of the inner shell before the air starts to ionize is E(a) = V m -1 What is the maximum possible charge on the inner capacitor? Solution: Q The electric field E(a) = Therefore the maximum charge is 4πε a Q max = 4πε E max (a)a = (3 16 V m -1 )(1 m) (9 1 9 N m C ) = C c) What is the maximum amount of energy stored in this capacitor? Solution: The energy stored is U max = Q max C = ( C) ()( 1 11 F) = J In the region between the two shells, the electric field energy density is given by Q u E = (1/ )ε E = (1/ )ε 4πε r = Q 3π ε r 4 To determine the total stored energy, integrate the energy density over a spherical shell of radius r, thickness dr, and volume 4πr dr, U E = u E dv = r=b r=a Q 3π ε r 4 4πr dr = Q 8πε r=b r=a dr r 1 8πε a 1 b = Q C = Q

4 Department of Physics: 8 d) When E(a) = V m -1 what is the potential difference between the shells? Solution: We can find the potential difference two different ways Using the definition of capacitance we have that ΔV = Q C = 4πε E(a)a (b a) 4πε ab ΔV = (3 16 V m -1 )(1 m)(1 m) ( m) We already calculated the potential difference in part a): Recall that E(a) = Q 4πε a or potential difference yielding = ΔV = Q 1 4πε a 1 b E(a)a(b a) b = V Q 4πε = E(a)a Substitute this into our expression for 1 ΔV = E(a)a a 1 b = (b a) E(a)a ab in agreement with our result above (b a) = E(a)a b

5 Department of Physics: 8 Problem Faraday s Law, Induced Electric Fields, and Lorentz Force Law An electron of the mass m e, and charge q = e is constrained to move in a circle of radius r by a time changing non-uniform magnetic field Assume that the magnet has cylindrical symmetry about the central axis passing through the poles and that the plane of the orbit is perpendicular to that symmetry axis Denote the magnetic field by B(r) = B z (r) ˆk where ˆk points form the north pole to the south pole Assume that z -component of the magnetic field, B z (r), varies as a function of the distance r from the symmetry axis, that it is symmetric about this axis, and that it is perpendicular to the plane of the orbit of the electron The magnitude of the average magnetic field over the electron s circular orbit is B ave = 1 πr B da Find a condition relating the rate of change in time of the average magnetic field db ave / dt, to the rate of change in time of the magnetic field, db / dt, in order for the electron to stay in a circular orbit Solution: Let s apply Faraday s Law to a circle of radius r between the poles of the magnet and shown in the figure below disk

6 Department of Physics: 8 Then Faraday s Law implies that circle E d s d 1 Eπr = πr dt πr = d dt disk disk B da, B da = πr db ave dt We can solve this equation for the electric field and find that E = r db ave ˆθ dt The force on the electron that is in a circular orbit or radius r with a tangential component of the velocity v θ, is then F e = e( E + v e B) = e r db ave dt ˆθ + v θ ˆθ Bz (r) ˆk = e r db ave dt ˆθ + v θ B z (r)ˆr, where B z (r) is the z -component of the magnetic field at radius z -component of the magnetic field r We now apply Newton s Second Law, F e = m e a, to the electron and obtain tangential and radial equations of motion er db ave dt ˆθ = m e dv θ dt ˆθ, ev θ B z (r)ˆr = (m e v θ / r)ˆr B z (r) = m e v θ er We can integrate the tangential equation and find that v θ = er m e B ave Then the z -component of the magnetic field becomes B z (r) = m e v θ er = m e er m e er B ave = B ave

7 Department of Physics: 8 Problem 3: RC Circuit (a) With the assumed directions of the current as shown, what does Kirchhoff s nd Law give you if you move clockwise around the lower loop (eg through the battery and R ) Be careful of your signs 1 I 1 R 1 +V = (b) With the assumed directions of the current as shown, what does Kirchhoff s nd Law give you if you move clockwise around the outer loop (eg through the battery, the capacitor, and R ) Be careful of your signs! What is the relation between I and Q(t)? Q C I R +V = The relation between I and Q(t) is dq / dt = I, ie, the rate at which the capacitor plate is charged is equal to the current across the resistor R (c) Just after the switch is closed what are the values of I 1, I, and Q(t = )? After the switch is closed, the currents are I 1 = V R 1, I = V R, and Q(t = ) =

8 Department of Physics: 8 (d) A long, long time after the switch is closed what are the values of I 1, I, and Q(t)? A long, long time after the switch is closed, I 1 = V R 1, I =, and Q = CV (e) Find expressions for I () t and Q(t) good for all t > Note that since R 1 is directly across the battery it is really just a distracter it in no way influences the current through the capacitor You can also see this in the equation from (b): = Q C I R +V = Q C R dq dt +V So we can just write down the solution for the time dependence of something that increases in time (the charge) and that decreases in time (the current) with time constant t = R C : Q(t) = Q f (1 e t/τ ) and I (t) = I e t/τ Q f just depends on the final voltage, which is the same as the voltage across the battery: Q f = CV The initial current is what you get when the capacitor is essentially a short: I = V R (f) The switch is now opened again after a long time How long after the switch is opened does it take the charge on the capacitor to fall to 1/ e of its value just before the switch was opened? After the switch is reopened again, the upper loops acts as a discharging RC -circuit, with R eq = R 1 + R (since the two resistors are connected in series) The time it takes for the charge to fall to 1/ e of its initial value is given by the time constant τ = R eq C = (R 1 + R )C

9 Department of Physics: 8 Problem 4 Consider a plane parallel capacitor of plate separation d and plate area A The capacitor is fully charged with charge Q on the positive plate Ignore edge effects Express your answers in terms of the given quantities a) Derive an expression for the magnitude of the electric field between the capacitor plates Draw the direction of the electric field on the sketch above Solution: We use Gauss s Law E da = q enc ε to find that EA cap = σ A cap ε Choose a unit vector k that points from the positively charged plate to the negatively charged plate Then the electric field between the plates is E = σ ε k = Q Aε k Note: we have assumed that there are no edge effects ie the electric field is zero outside the region between the plates b) What is the electric potential difference between the plates? Which plate is at a higher potential? Solution: The potential difference between the plates is given by

10 Department of Physics: 8 φ(+) φ( ) = + E dr = z = z = d The positive plate is at a higher potential c) What is the capacitance C? Solution: The capacitance is given by Q Aε k dz k = z = Q dz = Q d Aε Aε z = d C = Q φ(+) φ( ) = Q (Q / Aε )d = Aε d A current I flows counterclockwise as seen from above through each turn of a long air core cylindrical shaped solenoid that has N turns, length h, and radius a Ignore edge effects d) Derive an expression for the magnitude of the magnetic field inside the solenoid Draw the direction of the magnetic field on the sketch above Solution: We use Ampere s Law to calculate the magnetic field loop B dr = µ I through If we choose a rectangular loop of length l, then Ampere s Law becomes Bl = µ N h l I So the magnetic field inside the solenoid is given by B = µ N h I ˆk

11 Department of Physics: 8 where ˆk is a unit vector pointing upward e) What is the self-inductance L of the solenoid? Solution: The self-inductance of the solenoid is given by L = Φ total mag I = N Φ turn mag I = N I B da = NBπa I turn = N µ πa h The capacitor is now connected to the solenoid to form a resonant circuit without any resistance The capacitor is initially charged with charge Q on the positive plate, and when the switch is closed the current in the circuit is found to undergo sinusoidal oscillations f) What is the period T of oscillation of the charge on the capacitor plate? Solution: The period of oscillation is given by T = π ω = π LC = π N µ πa h Aε d = π Na µ ε g) A time interval Δ t = T / 4 has passed since the switch was closed What is energy stored in the capacitor? What is energy stored in the inductor? Solution: One-quarter cycle later the capacitor is uncharged and the current is maximum The energy in the capacitor is zero and the energy in the inductor is equal to the initial energy in the circuit, U mag = E = Q C = Q ε A d π A hd

12 Department of Physics: 8 Problem 5: Coaxial Cable and Power Flow A coaxial cable consists of two concentric long hollow cylinders of zero resistance; the inner has radius a, the outer has radius b, and the length of both is l, with l >> b, as shown in the figure The cable transmits DC power from a battery to a load The battery provides an electromotive force ε between the two conductors at one end of the cable, and the load is a resistance R connected between the two conductors at the other end of the cable A current I flows down the inner conductor and back up the outer one The battery charges the inner conductor to a charge Q and the outer conductor to a charge + Q (a) Find the direction and magnitude of the electric field E everywhere Answer: Consider a Gaussian surface in the form of a cylinder with radius r and length l, coaxial with the cylinders Inside the inner cylinder (r<a) and outside the outer cylinder (r>b) no charge is enclosed and hence the field is In between the two cylinders (a<r<b) the charge enclosed by the Gaussian surface is Q, the total flux through the Gaussian cylinder is Φ E = E da = E(π rl) Thus, Gauss s law leads to E(π rl) = q enc ε E = q enc π rl ˆr =, or Q ˆr (inward) for a < r < b, elsewhere πε rl (b) Find the direction and magnitude of the magnetic field B everywhere Answer: Just as with the E field, the enclosed current I enc in the Ampere s loop with radius r is zero inside the inner cylinder (r<a) and outside the outer cylinder (r>b) and hence the field there is In between the two cylinders (a<r<b) the current enclosed is I Applying Ampere s law, B d s = B(π r) = µ I enc, we obtain B = µ I πr ˆϕ (clockwise viewing from the left side) for a < r < b, elsewhere

13 Department of Physics: 8 (c) Calculate the Poynting vector S in the cable Answer: For a < r < b, the Poynting vector is S = 1 1 E B = µ µ On the other hand, for Q πε rl ˆr µ I πr ˆϕ = Q I 4π ε r ˆk (from right to left) l r < a and r > b, we have S = (d) By integrating S over appropriate surface, find the power that flows into the coaxial cable Answer: With da = ( π r dr) k ˆ, the power is P = S da = Q I b 1 4π ε l r (πrdr) = Q I a πε l ln b a S (e) How does your result in (d) compare to the power dissipated in the resistor? Answer: Because b Q Q b ε = E d s = dr = ln = IR a π rlε π lε a the charge Q is related to the resistance R by Q = πε lir The above expression for P ln(b / a) becomes P = πε lir I ln(b / a) πε l ln b a = I R which is equal to the rate of energy dissipation in a resistor with resistance R

14 Department of Physics: 8 Problem 6: Charging Capacitor A parallel-plate capacitor consists of two circular plates, each with radius R, separated by a distance d A steady current I is directed towards the lower plate and away from the upper plate, charging the plates a) What is the direction and magnitude of the electric field E between the plates? You may neglect any fringing fields due to edge effects Solution: If we ignore fringing fields then we can calculate the electric field using Gauss s Law, closed surface E d a = Q enc ε By superposition, the electric field is non-zero between the plates and zero everywhere else Choose a Gaussian cylinder passing through the lower plate with its end faces parallel to the plates Let A denote the area of the end face The surface charge density cap is given by σ = Q / π R Let ˆk denote the unit vector pointing from the lower plate to the upper plate Then Gauss Law becomes which we can solve for the electric field E A cap = σ A cap ε σ ˆ Q E = k = kˆ ε π R ε b) What is the total energy stored in the electric field of the capacitor?

15 Department of Physics: 8 Solution: The total energy stored in the electric field is given by 1 1 Uelec = ε E dv ε E π R d = volume Substitute the result for the electric field into the energy equation yields U elec Q Q d R d π π R ε π R ε 1 1 = ε = c) What is the time rate of change of the energy stored in the electric field? Solution: The rate of change of the stored electric energy is found by taking the time derivative of the energy equation d Qd dq U = elec dt π R ε dt The current flowing to the plate is equal to I dq = dt Substitute the expression for the current into the expression for the rate of change of the stored electric energy yields d QId Uelec dt = π R ε d) What is the magnitude of the magnetic field B at point P located between the plates at radius r < R (see figure above) As seen from above, is the direction of the magnetic field clockwise or counterclockwise Explain your answer Solution: We shall calculate the magnetic field by using the generalized Ampere s Law, closed path B d s = µ J da open surface + µ ε d dt open surface E d a We choose a circle of radius r < R passing through the point P as the Amperian loop and the disk defined by the circle as the open surface with the circle as its boundary We choose to circulate around the loop in the counterclockwise direction as seen from above This means that flux in the positive ˆk -direction is positive

16 Department of Physics: 8 The left hand side (LHS) of the generalized Ampere s Law becomes LHS = circle B d s = B πr The conduction current is zero passing through the disk, since no charges are moving between the plates There is an electric flux passing through the disk So the right hand side (RHS) of the generalized Ampere s Law becomes d d E RHS = µ ε E da = µ ε π r dt dt disk Take the time derivative of the expression for the electric field and the expression for the current, and substitute it into the RHS of the generalized Ampere s Law: d E µ Iπ r RHS = µ ε π = dt π R r Equating the two sides of the generalized Ampere s Law yields B µ Iπ r π r = π R Finally the magnetic field between the plates is then µ I ; π R B = r < r < R The sign of the magnetic field is positive therefore the magnetic field points in the counterclockwise direction (consistent with our sign convention for the integration direction for the circle) as seen from above Define the unit vector ˆθ such that is it tangent to the circle pointing in the counterclockwise direction, then B = µ I π R r ˆθ ; < r < R e) Make a sketch of the electric and magnetic field inside the capacitor

17 Department of Physics: 8 f) What is the direction and magnitude of the Pointing vector S at a distance r = R from the center of the capacitor? Solution: The Poynting vector at a distance r 1 S( r = R) = E B = R is given by µ r= R Substituting the electric field and the magnetic field (setting r = R ) into the above equation, and noting that ˆk ˆθ = ˆr, yields S(r = R) = 1 Q ˆk µ I µ π R ε π R So the Poynting vector points inward with magnitude Q ˆθ = π R ε Q I S ( r = R) = R π R ε π I π R ( ˆr) g) By integrating S over an appropriate surface, find the power that flows into the capacitor Solution: The power flowing into the capacitor is the closed surface integral P = closed surface S(r = R) da The Poynting vector points radially inward so the only contribution to this integral is from the cylindrical body of the capacitor The unit normal associated with the area vector for a closed surface integral always points outward, so on the cylindrical body

18 Department of Physics: 8 da = da rˆ Use this definition for the area element and the power is then Q I P = ( r = R) S da = ( ˆ) da ˆ R r r cylindrical body π R ε π cylindrical body The Poynting vector is constant and the area of the cylindrical body is π Rd P = Q I ( ˆr) da ˆr = Q π R ε π R π R ε cylindrical body The minus sign correspond to power flowing into the region, so I QId π Rd = π R π R ε h) How does your answer in part g) compare to your answer in part c)? Solution: The two expressions for power are equal so the power flowing in is equal to the change of energy stored in the electric fields

19 Department of Physics: 8 Problem 7: Superposition of Two Travelling Waves Suppose the electric field of an electromagnetic wave is given by the superposition of two waves E = E cos(kz ωt)î + E cos(kz + ωt)î You may find the following identities useful cos(kz ± ωt) = cos(kz)cos(ωt) sin(kz)sin(ωt) a) What is the associated magnetic field B(x, y,z,t)? Answer: We treat each contribution to the electric field separately Then we have that B = E c cos(kz ωt) ĵ E cos(kz + ωt) ĵ, c where we used that fact that that the amplitude of the magnetic field is related to the amplitude of the electric field by B = E / c, and the direction of the field satisfy dir( E B) = dir(propagation) The contribution E 1 = E cos(kz ωt)î the positive ˆk -direction, and the contribution E = E cos(kz + ωt)î negative ˆk -direction is propagating in is propagating in the b) Use the identity cos(a)sin(a) = (1/ )sin(a) to show that the energy per unit area per unit time (the Poynting vector) transported by this wave is S = (E / cµ )sin(kz)sin(ωt) ˆk Answer: We first use the identities cos(kz ± ωt) = cos(kz)cos(ωt) sin(kz)sin(ωt), to rewrite the electric field as E = E cos(kz ωt)î + E cos(kz + ωt)î = E cos(kz)cos(ωt)î The corresponding expression for the magnetic field is B = E c cos(kz ωt) ĵ E c cos(kz + ωt) ĵ = E c sin(kz)sin(ωt) ĵ The Poynting vector is then S= 1 1 E B = E µ o µ cos(kz)cos(ωt)î E o c sin(kz)sin(ωt) ĵ S= 4E cos(kz)cos(ωt)sin(kz)sin(ωt) ˆk cµ

20 Department of Physics: 8 Now use the identity cos(a)sin(a) = 1 sin(a) to rewrite the Poynting vector as S = E sin(kz)sin(ωt) ˆk cµ c) What is the time-average of the Poynting vector? Briefly explain your answer Answer: Note the time average of the Poynting vector is given by Therefore The integral is then 1 T S = 1 T = E 1 µ c T T T E S 1 T T Sdt µ c sin(kz)sin(ωt)dt ˆk T sin(kz) sin(ωt)dt ˆk sin(ωt)dt = 1 ωt cos(ωt) T =, where we used the fact that ωt = 4π and cos(ωt ) = cos(4π ) = 1 Thus S = This result may seem surprising at first If we rewrite the z -component of the Poynting vector as S z (t) = A(t)sin(kz), where A(t) = E µ c sin(ωt) is the time dependent amplitude Then the spatial dependence sin(kz) remains fixed with respect to time ie there is no propagation Only the amplitude A(t) varies sinusoidally with time In the graph below, we plot S z (t) = E sin(ωt)sin(kz) vs z for various values of µ c ωt =, ± π / 1, ± π / 6, ± π / 4, ± π / 3, ± π /

21 Department of Physics: 8 We see that the nodal structure does not change If we time average S z (t) we see that the sinusoidal amplitude averages to zero over one cycle

22 Department of Physics: 8 Problem 8 Measuring the Wavelength of Laser Light Suppose you shine a red laser through a pair of narrow slits, each of width a = 4µm, separated by a distance d = 5µm and allow the resulting interference pattern to fall on a screen a distance L = 4cm away This set up is shown in the figure to the right a) Will the center of the pattern (directly between the two holes) be an interference minimum or maximum? Answer: The center of the pattern will be a maximum because the waves from both slits travel the same distance to get to the center and hence are in phase (b) For the sizes given above, will these maxima be roughly equally spaced, or will they spread out away from the central peak? Answer: We get a maximum every time that the extra path length is an integral number of wavelengths: δ = d sinθ = mλ The spacing is the distance between these locations, y m+1 y m We can get y m from θ : sinθ m = y m L + y m = mλ d Let α m y m / L + y m Then α m = y m y m L + y L + y m m = α m y m ( 1 α m ) = α m L y m = α L m 1 α m α L 1 α m m

23 Department of Physics: 8 We have made the approximation that α m << 1, which is valid for the wavelengths and slit separations of this lab (it is order 1 3 ) As long as this approximation is valid, we can also ignore the term that goes like α m, and hence we find the maxima are equally spaced: y m+1 y m λl d (c) Approximately how many interference maxima will you see on one side of the pattern before their intensity is significantly reduced by diffraction due to the finite width a of the slit? Answer: The first single slit minimum appears at asinθ = λ So when we approach: m = d sinθ = d λ λ λ a = d a we will lose signal due to the diffraction minimum (d) Derive an equation for calculating the wavelength λ of the laser light from your measurement of the distance Δy between interference maxima Answer: Using what we derived for part b, Δy = y m+1 y m λl d λ = dδy L (e) In order to most accurately measure the distance between maxima Δy, it helps to have them as far apart as possible (Why?) Assuming that the slit parameters and light wavelength are fixed, what can we do in order to make Δy bigger? What are some reasons that can we not do this ad infinitum? Answer: We can increase the distance to the screen and measure the distance between distant interference maxima (eg m = 1 and m = 4), which increases distances, making them easier to measure, and then allows us to divide down any measurement errors Single Slit Interference Now that you have measured the wavelength λ of the light you are using, you will want to measure the width of some slits from their diffraction pattern When measuring diffraction patterns (as opposed to the interference patterns of problem 1) it is typically easiest to measure between diffraction minima

24 Department of Physics: 8 (f) Derive an equation for calculating the width a of a slit from your measurement of the distance Δy between diffraction minima Answer: Single slit minima obey the relationship asinθ = mλ, which is the same formula as two slit maxima So we can calculate the slit width from what we derived in part 1b (replacing the distance between the slits d with the width of the single slit a ): a = λl Δy (g) What is the width of the central maximum (the distance on the screen between the m = 1 and m = +1 minima)? How does this compare to the distance Δy between other adjacent minima? Answer: The central minimum is twice as wide as the distance between other minima It is: Δy central = λl a

25 Department of Physics: 8 Problem 8 Second-Order Bright Fringe A monochromatic light is incident on a single slit of width 8 mm, and a diffraction pattern is formed at a screen which is 8 m away from the slit The second-order bright fringe is at a distance 16 mm from the center of the central maximum What is the wavelength of the incident light? Solution: The general condition for destructive interference is λ y sin θ = m (1) a L where small-angle approximation has been made Thus, the position of the m-th order dark fringe measured from the central axis is y m λl = m () a Let the second bright fringe be located halfway between the second and the third dark fringes That is, The approximate wavelength of the incident light is then 1 1 λl 5λL yb = ( y + y3) = ( + 3) = (3) a a λ 3 3 a yb (8 1 m)(16 1 m) 7 = = 64 1 m (4) 5L 5(8 m)