course Phys 202 10:51 AM 10-26-09 028. `Query 28
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Given Solution: ** IL is the source. The law is basically an inverse square law and the angle theta between IL and the vector r from the source to the point also has an effect so that the field is B = k ' I L / r^2 * sin(theta). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery principles and general college physics problem 17.34: How much charge flows from each terminal of 7.00 microF capacitor when connected to 12.0 volt battery? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Q = CV Q = (7 * 10 ^ -6 F)(12 V) Q = 8.4 * 10 ^ -15 C confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Capacitance is stored charge per unit of voltage: C = Q / V. Thus the stored charge is Q = C * V, and the battery will have the effect of transferring charge of magnitude Q = C * V = 7.00 microF * 12.0 volts = 7.00 C / volt * 12.0 volts = 84.0 C of charge. This would be accomplished the the flow of 84.0 C of positive charge from the positive terminal, or a flow of -84.0 C of charge from the negative terminal. Conventional batteries in conventional circuits transfer negative charges. Explain how to obtain the magnetic field due to a circular loop at the center of the loop. ** For current running in a circular loop: Each small increment `dL of the loop is a source I `dL. The vector from `dL to the center of the loop has magnitude r, where r is the radius of the loop, and is perpendicular to the loop so the contribution of increment * `dL to the field is k ' I `dL / r^2 sin(90 deg) = I `dL / r^2, where r is the radius of the loop. The field is either upward or downward by the right-hand rule, depending on whether the current runs counterclockwise or clockwise, respectively. The field has this direction regardless of where the increment is located. The sum of the fields from all the increments therefore has magnitude B = sum(k ' I `dL / r^2), where the summation occurs around the entire loop. I and r are constants so the sum is B = k ' I / r^2 sum(`dL). The sum of all the length increments around the loop is the circumference 2 pi r of the loop so we have B = 2 pi r k ' I / r^2 = 2 pi k ' I / r. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK The solution doesn’t take into account that it is micro Farads. I think it should be multiplied by 10 ^ -6.
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Given Solution: STUDENT RESPONSE: We do have evidence that electric currents produce magnetic fields. This is observed in engineering when laying current carrying wires next to each other. The current carrying wires produce magnetic fields that may affect other wires or possibly metal objects that are near them. This is evident in the video experiment. When Dave placed the metal ball near the coil of wires and turned the generator to produce current in the wires the ball moved toward the coil. This means that there was an attraction toward the coil which in this case was a magnetic field. INSTRUCTOR COMMENT: Good observations. A very specific observation that should be included is that a compass placed over a conducting strip or wire initially oriented in the North-South direction will be deflected toward the East-West direction. ** How is the direction of an electric current related to the direction of the magnetic field that results? ** GOOD STUDENT RESPONSE: The direction of the magnetic field relative to the direction of the electric current can be described using the right hand rule. This means simply using your right hand as a model you hold it so that your thumb is extended and your four fingers are flexed as if you were holding a cylinder. In this model, your thumb represents the direction of the electric current in the wire and the flexed fingers represent the direction of the magnetic field due to the current. In the case of the experiment the wire was in a coil so the magnetic field goes through the hole in the middle in one direction. ** Query problem 17.35 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat would be the area of a .20 F capacitor if plates are separated by 2.2 mm of air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: C = e0 * A/d 0.20 F = (8.85 * 10 ^ -12 C^2 / N * m^2) * (A/ 2.2 * 10 ^ -3 m) A = 4.97 * 10 ^ 7 m^2 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If a parallel-plate capacitor holds charge Q and the plates are separated by air, which has dielectric constant very close to 1, then the electric field between the plates is E = 4 pi k sigma = 4 pi k Q / A, obtained by applying Gauss' Law to each plate. The voltage between the plates is therefore V = E * d, where d is the separation. The capacitance is C = Q / V = Q / (4 pi k Q / A * d) = A / (4 pi k d). Solving this formula C = A / (4 pi k d) for A for the area we get A = 4 pi k d * C If capacitance is C = .20 F and the plates are separated by 2.2 mm = .0022 m we therefore have A = 4 pi k d * C = 4 pi * 9 * 10^9 N m^2 / C^2 * .20 C / V * .0022 m = 5 * 10^7 N m^2 / C^2 * C / ( J / C) * m = 5 * 10^7 N m^2 / (N m) * m = 5 * 10^7 m^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 17.50 charge Q on capacitor; separation halved as dielectric with const K inserted; by what factor does the energy storage change? Compare the new electric field with the old. Note that the problem in the latest version of the text doubles rather than halves the separation. The solution for the halved separation, given here, should help you assess whether your solution was correct, and if not should help you construct a detailed and valid self-critique. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: PE = Q^2 / 2C and C = e0 * (A/d) If d is doubled to 2d: C = e0 * (A/2d) so PE = Q^2 / 2(e0 * (A/2d)) = Q^2 / 2(1/2 * e0 *A/d) The 2 and the ½ cancel each other making PE double confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For a capacitor we know the following: Electric field is independent of separation, as long as we don't have some huge separation. Voltage is work / unit charge to move from one plate to the other, which is force / unit charge * distance between plates, or electric field * distance. That is, V = E * d. Capacitance is Q / V, ratio of charge to voltage. Energy stored is .5 Q^2 / C, which is just the work required to move charge Q across the plates with the 'average' voltage .5 Q / C (also obtained by integrating `dW = `dQ * V = `dq * q / C from q=0 to q = Q). The dielectric increases capacitance by reducing the electric field, which thereby reduces the voltage between plates. The electric field will be 1 / k times as great, meaning 1/k times the voltage at any given separation. C will increase by factor k due to dielectric, and will also increase by factor 2 due to halving of the distance. This is because the electric field is independent of the distance between plates, so halving the distance will halve the voltage between the plates. Since C = Q / V, this halving of the denominator will double C. Thus the capacitance increases by factor 2 k, which will decrease .5 Q^2 / C, the energy stored, by factor 2 k. ** STUDENT QUESTION: I am very confused on the correct answer. I assumed the voltage would stay constant. However, I think what the true answer is that voltage will increase by 1/k? My final answer is the same as yours, that the energy will increase by 2k. INSTRUCTOR RESPONSE: The capacitor is already charged, so Q remains constant. The effect of the dielectric is to decrease the electric field, which by itself would decrease the voltage to 1/k of its former value, which increases the capacitance by factor k. The effect of halving the distance is to decrease the voltage by another factor of 2, which increases the capacitance by factor 2. Since Q remains constant, the energy .5 Q^2 / C decreases by factor 2 k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Ok – I just showed the equations instead of talking through the theory. ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery univ 24.50 (25.36 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery. What is the capacitance of this capacitor? ********************************************* Question: `qquery univ 24.68 (25.52 10th edition). Solid conducting sphere radius R charge Q.