Query28Phys

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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

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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): ------------------------------------------------ Self-critique 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:OK

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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

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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:OK

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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:OK

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Given Solution: ** Fundamental principles include the fact that the electric field is very neary constant between parallel plates, the voltage is equal to field * separation, electric field from a single plate is 2 pi k sigma, the work required to displace a charge is equal to charge * ave voltage, and capacitance is charge / voltage. Using these principles we reason out the problem as follows: If the 4.7 mm separation experiences a 12 V potential difference then the electric field is E = 12 V / (4.7 mm) = 12 V / (.0047 m) = 2550 V / m, approx. Since the electric field of a plane charge distribution with density sigma is 2 pi k sigma, and since the electric field is created by two plates with equal opposite charge density, the field of the capacitor is 4 pi k sigma. So we have 4 pi k sigma = 2250 V / m and sigma = 2250 V / m / (4 pi k) = 2250 V / m / (4 pi * 9 * 10^9 N m^2 / C^2) = 2.25 * 10^-8 C / m^2. The area of the plate is .0256 m^2 so the charge on a plate is .0256 m^2 * 2.25 * 10^-8 C / m^2 = 5.76 * 10^-10 C. The capacitance is C = Q / V = 5.67 * 10^-10 C / (12 V) = 4.7 * 10^-11 C / V = 4.7 * 10^-11 Farads. The energy stored in the capacitor is equal to the work required to move this charge from one plate to another, starting with an initially uncharged capacitor. The work to move a charge Q across an average potential difference Vave is Vave * Q. Since the voltage across the capacitor increases linearly with charge the average voltage is half the final voltage, so we have vAve = V / 2, with V = 12 V. So the energy is energy = vAve * Q = 12 V / 2 * (5.76 * 10^-10 C) = 3.4 * 10^-9 V / m * C. Since the unit V / m * C is the same as J / C * C = J, we see that the energy is 3.4 * 10^-9 J. Pulling the plates twice as far apart while maintaining the same voltage would cut the electric field in half (the voltage is the same but charge has to move twice as far). This would imply half the charge density, half the charge and therefore half the capacitance. Since we are moving only half the charge through the same average potential difference we use only 1/2 the energy. Note that the work to move charge `dq across the capacitor when the charge on the capacitor is `dq * V = `dq * (q / C), so to obtain the work required to charge the capacitor we integrate q / C with respect to q from q = 0 to q = Q, where Q is the final charge. The antiderivative is q^2 / ( 2 C ) so the definite integral is Q^2 / ( 2 C). This is the same result obtained using average voltage and charge, which yields V / 2 * Q = (Q / C) / 2 * Q = Q^2 / (2 C) Integration is necessary for cylindrical and spherical capacitors and other capacitors which are not in a parallel-plate configuration. ** query univ 24.49/24.51 (25.37 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery. If the battery remains connected and plates are pulled to separation 9.4 mm then what are the capacitance, charge of each plate, electric field, energy stored? The potential difference between the plates is originally 12 volts. 12 volts over a 4.7 mm separation implies electric field = potential gradient = 12 V / (.0047 m) = 2500 J / m = 2500 N / C, approx.. The electric field is E = 4 pi k * sigma = 4 pi k * Q / A so we have Q = E * A / ( 4 pi k) = 2500 N / C * (.16 m)^2 / (4 * pi * 9 * 10^9 N m^2 / C^2) = 5.7 * 10^-7 N / C * m^2 / (N m^2 / C^2) = 5.7 * 10^-10 C, approx.. The energy stored is E = 1/2 Q V = 1/2 * 5.6 * 10^-10 C * 12 J / C = 3.36 * 10^-9 J. If the battery remains connected then if the plate separation is doubled the voltage will remain the same, while the potential gradient and hence the field will be halved. This will halve the charge on the plates, giving us half the capacitance. So we end up with a charge of about 2.8 * 10^-10 C, and a field of about 1250 N / C. The energy stored will also be halved, since V remains the same but Q is halved. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK

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Given Solution: ** We will find the energy density function then integrate that density function over all of space to find the total energy of the distribution. We will compare this with the energy required to assemble the distribution, and will find that the two are equal. To integrate the energy density over all space we will find the total energy in a thin spherical shell of radius r and thickness `dr, then use this result to obtain our integral. Then we will integrate to find the work required to assemble the charge on the surface of the sphere. Energy density, defined by dividing the energy .5 C V^2 required to charge a parallel-plate capacitor by the volume occupied by its electric field, is Energy density = .5 C V^2 / (volume) = .5 C V^2 / (d * A), where d is the separation of the plates and A the area of the plates. Since C = epsilon0 A / d and V = E * d this gives us .5 epsilon0 A / d * (E * d)^2 / (d * A) = .5 epsilon0 E^2 so that Energy density = .5 epsilon0 E^2, or in terms of k = 4 pi / epsilon0 Energy density = 1 / (8 pi k) E^2. Since your text uses epsilon0 I'll do the same on this problem. In this problem the epsilon0 notation makes a good deal of sense: For the charged sphere the electric field for r < R is zero, since there can be no electric field inside a conductor. For r > R the electric field is E = Q / (4 pi epsilon0 r^2), and therefore energy density = .5 epsilon0 E^2 = .5 epsilon0 Q^2 / (16 pi^2 epsilon0^2 r^4) = Q^2 / (32 pi^2 epsilon0 r^4). The energy density (i.e., the energy per unit of volume) between r and r + `dr is nearly constant if `dr is small. As we saw above the energy density will be approximately Q^2 / (32 pi^2 epsilon0 r^4). The volume of space between r and r + `dr is approximately A * `dr = 4 pi r^2 `dr. The expression for the energy lying in the shell between distance r and r + `dr is therefore approximately energy in shell = energy density * volume = Q^2 / (32 pi^2 epsilon0 r^4) * 4 pi r^2 `dr = Q^2 / (8 pi epsilon0 r^2) `dr. This leads to a Riemann sum over radius r. As we let `dr approach zero the Riemann sum approaches an integral with integrand Q^2 / (8 pi epsilon0 r^2), integrated with respect to r. To get the energy between two radii we therefore integrate this expression between those two radii. If we integrate this expression from r = R to infinity we get the total energy of the field of the charged sphere. This integral gives us total energy = Q^2 / (8 pi epsilon0 R) (alternatively using k = 4 pi k / epsilon0, this result would be k Q^2 / (2 R). This form is less complicated to the eye and will be used in the comparison below.) We compare this with the work required to charge the sphere: To bring a charge `dq from infinity to a sphere containing charge q requires work k q / R `dq To charge the entire sphere the work would therefore be the integral of k q / R with respect to q. We integrate from q = 0 to q = Q, obtaining the total work required to charge the sphere. Our antiderivative is k (q^2 / 2) / r. If we evaluate this antiderivative at lower limit 0 and upper limit Q we get the total work, which is k Q^2 / (2 R). This agrees with our previous result, obtained by integrating the energy density of the field. Since k = 1 / (4 pi epsilon0) the work is k Q^2 / (2 R) = Q^2 / (8 pi epsilon0 R). So the energy in the field is equal to the work required to assemble the charge distribution. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK"

Query28Phys

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&#This looks good. Let me know if you have any questions. &#