Physics 9 Spring 2011 Midterm 1 Solutions

Transcription

1 Physics 9 Spring 2011 Midterm 1 s For the midterm, you may use one sheet of notes with whatever you want to put on it, front and back. Please sit every other seat, and please don t cheat! If something isn t clear, please ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR FINAL ANSWERS! You have the full length of the class. If you attach any additional scratch work, then make sure that your name is on every sheet of your work. Good luck! 1. The electric potential in a region of space is V = 200 x2 + y 2 V, where x and y are in meters. (a Determine the electric field, E, in terms of the unit vectors î and ĵ. (b What are the magnitude and direction of the electric field at (x, y = (2.0 m, 2.0 m? Give the direction as an angle cw or ccw (specify which from the positive x axis. (a For a potential of more than one variable, V (x, y, then ( V E = V = x î + V y ĵ. Now, V = 200x, while V =. Thus, the electric field is x (x 2 +y 2 3/2 y (x 2 +y 2 3/2 200y ( xî + yĵ E (x, y = 200 (x 2 + y 2. 3/2 (b Now, at (x, y = (2, 2, then ( 2î + 2ĵ E (2, 1 = 200 (î ( = ĵ, 3/2 which has a magnitude E = = 25 V/m. The angle, measured counterclockwise from the x axis is θ = tan 1 ( Ey E x = tan 1 (1 = 45. 1

2 2. An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V 0, and the electric field outside the cylinder is E r = λ 2πɛ 0 r. Find the potential relative to the surface at a point that is distance r from the axis, assuming r > R. For a given electric field, the change in potential is V = rf r i E d s, where V = V (r f V (r i. With our electric field we have, after taking r f = r, and r i = R, R λ dr V (r V (R = r 2πɛ 0 r = λ ( r ln. 2πɛ 0 R But, V (R is the potential when r = R; in other words, it is the potential on the surface of the sphere, which we ve been told is V 0. So, we find that V (r = V 0 λ ( r ln. 2πɛ 0 R 2

3 3. An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge +Ze (the nucleus at the center of a sphere of radius R with uniformly distributed negative charge Ze. Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere. (a Show that the electric field inside this atom is E in = Ze 4πɛ 0 ( 1 r 2 r R 3 (b What is E at the surface of the atom? Is this the expected value? Explain. (c A uranium atom has Z = 92 and R = 0.10 nm. What is the electric field strength at r = 1 2 R?. The atom is seen in the figure to the right. It contains a positively charged nucleus of charge +Ze, which is embedded inside a uniform sphere of charge Ze. We are interested in the field inside the sphere, so we take a spherical gaussian surface of radius r. (a Since the electric field is radial, and everywhere constant on the Gaussian surface, Gauss s law gives the usual result that E da = EA = E (4πr 2 = Q encl ɛ 0. Now, what s Q encl? This problem is very similar to the last one, except with the addition of +Ze at the center. So, the total enclosed charge is Q encl = +Ze negative charge. ( The negative charge is just the charge density, times the enclosed volume, 4π 3 r3 = Ze r3. Thus, R 3 Ze 4 3 πr3 Q encl = Ze (1 r3, R 3 which, using Gauss s law as discussed above, gives E in = Ze ( 1 4πɛ 0 r r. 2 R 3 (b On the surface of the atom, r = R, and so the electric field is zero. This is to be expected, since on the surface the enclosed charge is zero (+Ze Ze, and so the electric flux is zero from Gauss s law. The atom looks neutral outside the surface. 3

4 (c When r = R/2, then E ( R = Ze ( 4 2 4πɛ 0 R 1 2 2R 2 = 7 Ze 2 4πɛ 0 R. 2 Plugging in the numbers gives ( R E = 7 Ze 2 2 4πɛ 0 R = ( = N/C. So, E = N/C, which is a very strong electric field! 4

5 4. Consider the circuit in the diagram to the right. (a Determine the change in voltage through each resistor. (b How much power is dissipated by each resistor? First, let s figure out the current in the circuit. Since the two resistors are in series, we can replace them by an equivalent resistance, R eqv = R 1 + R 2 using Kirchhoff s loop law, Vi = E IR 1 IR 2 = 0. This gives for the current, I = E R 1 +R 2 = 12 = 12 = = 0.40 amp. (a The change in voltage through each resistor is V = IR, and so V 1 = IR 1 = = 4.8 V V 2 = IR 2 = = 7.2 V. (b Now, the power dissipated by each resistor, R i, is P i = I V i = I 2 R i. So, we find for each resistor P 1 = I 2 R 1 = ( = 1.92 W P 2 = I 2 R 2 = ( = 2.9 W. 5

6 The potential energy between a pair of neutral atoms or molecules is very well-approximated by the Lennard-Jones Potential, given by the expression [ (σ 12 ( σ ] 6 P E(r = 4ɛ, r r where ɛ and σ are constants, and r is the distance between the molecules. The potential energy is plotted in the figure to the right. The vertical axis is in units of ɛ, while the horizontal axis is in units of σ. Extra Credit Question!! The following is worth 10 extra credit points! Energy Molcular Bond Energy Distance (a Why does the potential energy approach zero as the distance gets bigger? (b At what separation distance, in terms of σ and ɛ, is the potential energy zero? (c At approximately what distance is the system in equilibrium? What is the potential energy at that distance? (Express your answers in terms of σ and ɛ. (d How much energy would you need to add to the system at equilibrium in order to break the molecular bonds holding it together? Why? (e How much energy is released in the breaking of those molecular bonds? Why? Note - no calculation is needed to answer these problems! (a As the two molecules get further apart, the attractive force between them gets weaker and weaker. When the are very far apart, they hardly interact at all - they are basically free molecules. The potential energy of a free particle is zero, since potential energy depends on the interaction between multiple particles. (b We can just read the value off from the graph. We see that the potential energy crosses the x axis when x = 1, which means that r = σ. We can see this from the equation, too: setting r = σ gives P E(σ = 0. 6

7 (c The system is in equilibrium when the net force on it is zero. Since the force is the slope of the potential energy graph, this happens when the slope is zero. The potential energy graph has zero slope when it s at it s minimum point. Checking the graph, we see that this happens right around x 1.15, or r 1.15σ. We could check the exact answer by finding d dr (P E(r = 0, which gives r = 21/6 σ 1.12σ, and so we were close on our guess. The energy at this distance can just be read off the graph, giving y = 3, or P E = 3ɛ. (d In order to break the molecular bonds apart, we d need to raise the energy to zero. At equilibrium the energy is P E = 3ɛ, and so we d need to add +3ɛ units of energy. (e There is no energy released in breaking these molecular bonds - we had to add the energy to break these bonds. Energy is never released in the breaking of bonds! One can obtain energy by breaking a less stable bond, then forming a more stable bond. The more stable bond has a more negative potential energy (a deeper potential well. The difference in energy between the initial and final states is released to the environment. This is where the energy comes from in the ATP reactions, and not by releasing energy from the breaking of bonds! 7

8 Some Useful Constants. Some Possibly Useful Information Coulomb s Law constant k 1 4πɛ 0 = Nm2 C 2. The magnetic permeability constant µ 0 = 4π 10 7 N A 2. Speed of Light c = m/s. 11 Nm2 Newton s Gravitational Constant G = kg 2 The charge on the proton e = C The mass of the electron, m e = kg. The mass of the proton, m p = kg. Boltzmann s constant, k B = J/K. 1 ev = Joules 1 MeV = 10 6 ev. 1 Å = meters. Planck s constant, h = J s = ev s. The reduced Planck s constant, h 2π = J s = ev s. Some Useful Mathematical Ideas. { x n+1 x n n 1, n+1 dx = ln (x n = 1. dx a = ln ( x + a 2 + x x 2 x dx a = a 2 + x x 2 Other Useful Stuff. The force on an object moving in a circle is F = mv2 r. Kinetmatic equations x(t = x 0 + v 0x t a xt 2, y(t = y 0 + v 0y t a yt 2. The binomial expansion, (1 + x n 1 + nx, if x 1. 8