SOLUTIOJN TO EXAM 1 - Exam 1 solutions Name: Physics 204...


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Unformatted text preview: Exam 1 solutions Name: Physics 204 morning summer 2008 Instructions: Score • The responsibility is yours to show me that you understand what you are doing so please show all your work. If you feel you are stuck on a particular part explain as much of your thinking as you can so I can give you the maximum amount of partial credit (explain the strategy you are using, draw pictures/graphs, etc.) 1: • Use the backs of the exam pages if you need extra space. 4: • Each problem is worth 10 points. The problems are not all of the same difficulty though, so budget your time wisely. 5: • The last pages of the exam contain useful physical constants, conversion factors, and other data. Please check there first if you feel you need more information than is given in the problem. 2: 3: Total /50 Good luck! 1. Three glass spheres of positive charge Q = 10 −9 C are arranged so that they are located at the corners of a square of side L = 0.1m . A proton is placed at the center of the square, then released so it may move freely. a. Sketch a picture of the situation and describe in words the resulting motion of the proton. Be as specific as you can with your description. b. Predict the speed of the proton once it once it reaches the empty corner. (Hint: Remember to use the principle of superposition.) The proton will experience a non‐constant force in the direction directly towards the empty corner since the electric field is non‐ uniform. To predict the speed of the proton when it reaches the empty corner I’ll use energy principles. Since this is an isolated system the total energy of the system is a conserved quantity. Ei = E f Qq ⎞ ⎛ Qq 1 = mpv 2 + ⎜ 2k p + k p ⎟ L 2 2 2L ⎠ ⎝ Qqp Qq Qq Qq ⎛ 1 1 ⎞ ⇒ mpv 2 = 3k − 2k p − k p = k p ⎜ 3 2 − 2 − ⎟ 2 L L ⎝ 2L 2⎠ L 2 ⇒v = k Qqp mpL (5 2 −4 2L ) L p+ L Qqp L + + ⇒ K i + UE ,i = K f + UE , f ⇒ 0 + 3k + ⇒v = k (5 mL Qqp ) (9 × 10 Nm 2 −4 = 9 p 2 (10 C )(1.6 × 10 C ) ) 1.67 × 10 kg 0.1m ( 5 ( )( ) −9 C 2 −19 −27 ) 2 − 4 = 1.63 × 105 m s 2. Two glass spheres (radius 3cm) are attached to opposite ends of a plastic rod (30cm long). −8 The spheres are rubbed with silk making them positively charged (charge 2.0 × 10 C ). Imagine a line passing through both spheres. Now, consider a point on that line 50m away from the center of the rod. a. Since the point you are interested in is very far from the spheres compared to the distance between them you can approximate the entire system of rod and 2 spheres as a single point‐like particle. Predict the value of the electric field at the point of interest. b. Now, consider a point only 1m away from the center of the rod. Using the same approximation (even though it’s not that reasonable anymore) you used in part a. predict the electric field at this new point. c. Then, predict the electric field at this new point exactly, without using the approximation. d. Lastly, calculate the percent difference between the approximate value and the exact value. This is a measure of how reasonable the approximation is. Since 50m is so much greater than 30cm, the electric field created by the system will be approximately the same as that of a point charge. I’ll choose a coordinate system where the origin is located between the two spheres and the x axis is perpendicular to the rod. −8 2q 9 2 2 4.0 × 10 C E = k 2 = ( 9 × 10 Nm C ) = 0.144 N C 2 r ( 50m ) The electric field points in the +x direction. Using the same approximation 1m away (inappropriate at such a close distance though) gives E =k 2q 4.0 × 10 −8 C = ( 9 × 109 Nm2 C 2 ) = 360 N C 2 r2 (1m ) The electric field points in the E +x direction. To determine the electric field at 1m exactly I need to add the r a x contributions from the two spheres as vectors. However, the electric field contributions from each sphere both point in the +x direction, so they can be added directly. ⎛ ⎞ 1 1 = kq ⎜ + ⎟ 2 2 ⎜ ( x + a + r )2 ( x − a − r )2 ⎟ (x +a + r) (x −a −r) ⎝ ⎠ ⎛ ⎞ 1 1 ⇒ E = ( 9 × 109 Nm2 C 2 )( 2.0 × 10 −8 C ) ⎜ + ⎟ ⎜ (1m + 0.15m + 0.03m )2 (1m − 0.15m − 0.03m )2 ⎟ ⎝ ⎠ ⇒ E = 397 N C E =k q +k q The approximate value differs from the exact value by 397 N C − 360 N C × 100% = 9.32% 397 N C 3. You and your friend are studying for the next physics exam. Your friend shows you a diagram of a circuit with a single battery and a single light bulb. He then adds a second identical light bulb in parallel with the first and says “by doing this the brightness of the first light bulb will be halved because there are now two light bulbs in the circuit. In fact, the second light bulb will also have this same brightness.” You are not so sure… Maybe your friend is right, but maybe he isn’t. a. What, if anything, will happen to the brightness of the first bulb? What will be the brightness of the second bulb? Explain your reasoning in detail. b. What, if anything, will happen to the brightness of the first bulb if instead the second light bulb were added in series? What will be the brightness of the second bulb? Explain your reasoning in detail. The brightness of the first bulb will not change, and the second bulb will be just as bright as the first. Since the second light bulb is added in parallel each light bulb will receive the full potential difference of the battery. Since the resistance of the light bulb remains the same as well the power output ( P = ΔV 2 R ) will not change. If the second light bulb is added in series however the potential difference across each light bulb will be half the potential difference across the battery. This will result in both light bulbs being 1/4 as bright as when a single bulb was in the circuit. 4. Consider the circuit shown. ε = 120V R1 R1 = 150Ω R2 = 75Ω R3 = 75Ω R3 ε R2 a. Determine the power output of each light bulb (shown as resistors). b. What assumptions did you make about the circuit? c. If this circuit were in operation in a significantly colder environment, would your answers to part a. increase, decrease, or remain the same. Explain your reasoning. First, I’ll simplify the circuit and find the current flowing through the power supply (and therefore through resistor 2.) Resistors 1 and 3 are in parallel and resistor 2 is in series with them. So the equivalent resistance of the circuit is −1 −1 ⎛1 1⎞ 1 ⎞ ⎛ 1 + Req = R2 + ⎜ + ⎟ = 75Ω + ⎜ ⎟ = 125Ω ⎝ 150Ω 75Ω ⎠ ⎝ R1 R3 ⎠ The current through the power source (and therefore resistor 2) is I2 = ε Req = 120V = 0.96 A 125Ω The power output of this light bulb is P2 = I22R2 = ( 0.96 A ) ( 75Ω ) = 69.1W 2 To find the current through the other two resistors I need to find the potential difference across them first. Each will have a potential difference of ε − I2R2 = 120V − ( 0.96 A )( 75Ω ) = 48V The power outputs of resistors 1 and 3 are then ΔV 2 ( 48V ) P1 = 1 = = 15.4W 150Ω R1 2 ΔV 2 ( 48V ) P3 = 3 = = 30.7W 75Ω R3 2 I assumed that neither the power supply nor the wires had any resistance. If the circuit were operating in a colder environment the resistance of all the resistors would be lower. This would result in larger currents flowing in the circuit. Since P = I R the increase in current is more significant than the decrease in resistance, so all of the light bulbs will be brighter. 2 5. A very small and light aluminum ball is suspended by an insulating thread between two electrically neutral aluminum plates. Each plate is mounted vertically on an insulating plastic base. Then, a power supply is connected to the plate on the left which keeps it at a constant positive charge. As soon as this happens the ball swings all by itself toward the left plate. As soon as the ball touches the plate it bounces off it. Then, the ball swings across to the other plate, and bounces off that one. Then it swings back across to the left plate, and so on. This bouncing between the two plates happens several dozen times, then starts slowing down. Eventually the bouncing stops and the ball comes to rest hanging in between the plates again. Explain in detail why this is happening. Be very specific, and use diagrams to help you in your explanation. Once the power supply causes the left plate to become positive, the ball polarizes and is attracted to it. Once the ball touches the left plate electrons are transferred from the ball to the plate. The ball is repelled by the left plate and swings toward the right plate. Since the ball is positive and the right plate is neutral the ball is attracted to the right plate. When the ball touches the right plate electrons are transferred to the ball nearly neutralizing it. Since the ball is nearly neutral it will be attracted by both plates, but it will be attracted to the left plate more since the left plate is (at the moment) more positively charged. So, the ball swings back to the left plate, touches it, and the whole sequence of events repeats. Eventually, so many electrons have been transferred from the right plate to the left plate by the ball that the forces being exerted on the ball will balance and it will remain hanging in the middle. k= ...
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