#$&* course PHY 202 028. `Query 28 was submitted 22 July 2011 around 10:15 AM. 028. `Query 28
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Given Solution: ** IL is the source. The law is basically an inverse square law and the angle theta between IL and the vector r from the source to the point also has an effect so that the field is B = k ' I L / r^2 * sin(theta). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery principles and general college physics problem 17.34: How much charge flows from each terminal of 7.00 microF capacitor when connected to 12.0 volt battery? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Capacitance is stored charge per unit of voltage: C = Q / V. The charge Q acquired by each plate of a capacitor is given by the equation Q = C * V. Here, C is the capacitance in farad and V is the potential difference in volts. The battery will have the effect of transferring charge of magnitude: Q = C * V Q = 7.00 microF * 12.0 volts Q = 7.00 microC / volt * 12.0 volts Q = 84.0 microC of charge. This would be accomplished the flow of 84.0 microC of positive charge from the positive terminal, or a flow of -84.0 microC of charge from the negative terminal. Conventional batteries in conventional circuits transfer negative charges. Therefore, a charge of 84.0 microC flow from each terminal of the battery. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 microC / volt * 12.0 volts = 84.0 microC of charge. This would be accomplished the the flow of 84.0 microC of positive charge from the positive terminal, or a flow of -84.0 microC of charge from the negative terminal. Conventional batteries in conventional circuits transfer negative charges. STUDENT COMMENT Ok. I didn’t really understand the +/- explanation though. INSTRUCTOR RESPONSE The positive terminal of a battery attracts negative charges, and/or repels positive charges. The negative terminal attracts positive charges, and/or repels negative charges. In a circuit where the available 'free charges' are negative, as in most circuits consisting of metal wires and various circuit elements. In such a circuit negative charges that reach the positive terminal are 'pumped' through the battery to the negative terminal (they wouldn't go there naturally; it takes energy to pull them away from the positive and get them to move to the negative terminal), where they are repelled. The result is a flow of negative charges toward the positive terminal, then away from the negative. This is completely equivalent to what would happen if the charge carriers were positive, moving in the opposite direction, away from the positive terminal and toward the negative. For a good time after circuits were put into use, nobody knew whether the charge carriers were positive or negative, or perhaps a mix of both. The convention prior to that time was that the direction of the current was the direction in which positive charge carriers would move (away from positive terminal, torward the negative). By the time the nature of the charges was discovered, the textbooks and engineering manuals had been around for awhile, and there was no way to change them. So the convention continues. STUDENT QUESTION I see that the unit is C not Farad? I understand the volt canceling out, but I thought capacity was measured 1 C/V? INSTRUCTOR RESPONSE Good question. The problem asked for the amount of charge. Charge is measured in Coulombs, abbreviated C. It's easy to confuse the C that stands for capacitance with the C that stands for Coulombs: • The unit C stands for Coulombs. • The variable C stands for capacitance. • To avoid confusion we have to be careful to keep the context straight. A Farad is a Coulomb per volt (C / V). So the unit of capacitance C is the Farad, or (C / V), where the C in the units stands for Coulombs. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: Explain how to obtain the magnetic field due to a circular loop at the center of the loop. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: To obtain the magnetic field due to a circular loop at the center of the loop, the loop of radius r can be thought of as a series of very short segments, of total length 2 `pi r equal to the circumference of the circle. • The segments can be made as short as desired, so the approximation to each field can be made as accurate as desired. • Each segment `dL is a source I `dL and is perpendicular to a line from the center of the segment to the center of the circle., • Each segment thus contributes `dB = k ' I `dL / r^2 to the field at the center. • All contributions are in the same direction, as can be easily verified. 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 = k ' I `dL / r^2 sin(90 deg) = I `dL / r^2. 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. When all the magnetic field contributions are added, we obtain • B = `sum(`dB) = `sum( k ' I `dL / r^2 ). Since k ' , I and r are identical for all contributions, we obtain • B = `sum ( k ' I dL / r^2 ) = k ' I / r^2 * `sum(`dL). Since the sum of all the `dL contributions for one loop is just the circumference 2 `pi r of the loop, we finally have • B = k ' I / r^2 * 2 `pi r = 2 `pi k ' I / r. If there are N loops, then this result is multiplied by N. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery magnetic fields produced by electric currents. What evidence do we have that electric currents produce magnetic fields? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 3 ********************************************* Question: `qWhat would be the area of a .20 F capacitor if plates are separated by 2.2 mm of air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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 given 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: 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 A = 4 pi * 9 * 10^9 N m^2 / C^2 * .20 C / V * .0022 m A = 5 * 10^7 N m^2 / C^2 * C / ( J / C) * m A = 5 * 10^7 N m^2 / (N m) * m A = 5 * 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. ** STUDENT QUESTION I am not seeing where the 4pi k d came from... INSTRUCTOR RESPONSE 4 pi k Q / A * d is the same as 4 pi k d Q / A, by order of operations. So Q / (4 pi k Q / A * d) simplifies to A / (4 pi k d). The electric field near the surface of of a flat plate is 2 pi k * Q / A, as we find using the flux picture (total flux of charge Q is 4 pi k Q; a rectangular Gaussian surface and symmetry arguments are used to show that half the flux exits each end of the surface, resulting in field 2 pi k Q / A). The electric field between two oppositely charged plates is therefore 4 pi k * Q / A. Multiplying this field by the distance d between plates gives us the voltage. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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 self-critique. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For a capacitor, the electric field between the plates is 4 pi k Q / A, as long the separation d of the plates is small compared to the dimensions of the plates, and is independent of the separation. Voltage is work / unit charge to move from one plate to the other. Since work = force * distance, work / unit charge is which is force / unit charge * distance between plates. Equivalently, since the electric field is the force per unit charge, work / unit charge is electric field * distance. • V = E * d Capacitance is Q / V, ratio of charge to voltage. Energy stored is ½ Q^2 / C, which is just the work required to move charge Q across the plates with the 'average' voltage ½ Q / C (also obtained by integrating `dW = `dQ * V = `dq * q / C from q=0 to q = Q). For the present situation we halve the separation of the plates and insert a dielectric with constant k. For a given Q, then, the electric field is fixed so that halving the separation d halves the voltage V = E * d. Halving the voltage V doubles the capacitance Q / V. Then inserting the dielectric reduces the field E, thereby reducing the voltage and increasing the capacitance by factor k. Thus the capacitance increases by factor 2 k. For given Q, this will decrease the stored energy ½ Q^2 / C by factor 2 k. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For a capacitor we know the following: • The electric field between the plates is 4 pi k Q / A (see solution to preceding problem), as long the separation d of the plates is small compared to the dimensions of the plates, and is independent of the separation. • Voltage is work / unit charge to move from one plate to the other. Since work = force * distance, work / unit charge is which is force / unit charge * distance between plates. Equivalently, since the electric field is the force per unit charge, work / unit charge is 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. For the present situation we halve the separation of the plates and insert a dielectric with constant k. For a given Q, then, the electric field is fixed so that halving the separation d halves the voltage V = E * d. Halving the voltage V doubles the capacitance Q / V. Then inserting the dielectric reduces the field E, thereby reducing the voltage and increasing the capacitance by factor k. Thus the capacitance increases by factor 2 k. For given Q, this will decrease the stored energy .5 Q^2 / C 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 ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!