course phy 232 Is my test proctoring ready to go? I filled out the proctor form a week ago and have received no responses. aeܓ}yxt[[assignment #006
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18:58:22 Query Principles and General Physics 17.4: work by field on proton from potential +135 V to potential -55 V.
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RESPONSE --> proton=1.60*10^-19 C V=J/c dV= Va-Vb dv=dW/c dv*q=Wd W= 3.04*10^-17 J
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19:01:20 The change in potential is final potential - initial potential = -55 V - (125 V) = -180 V, so the change in the potential energy of the proton is -180 V * 1.6 * 10^-19 C = -180 J / C * 1.6 * 10^-19 C = -2.9 * 10^-17 J. In the absence of dissipative forces this is equal and opposite to the change in the KE of the proton; i.e., the proton would gain 2.9 * 10^-17 J of kinetic energy. Change in potential energy is equal and opposite to the work done by the field on the charge, so the field does 2.9 * 10^-17 J of work on the charge. Since the charge of the proton is equal in magnitude to that of an electron, he work in electron volts would be 180 volts * charge of 1 electron= 180 eV.
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RESPONSE --> The 180 should be 190 because it was 135V initially but the protaon potential energy should be negative because the V decreases to -55v.
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19:11:12 Query Principles and General Physics 17.8: Potential difference required to give He nucleus 65.0 keV of KE.
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RESPONSE --> The problem wants to get 65000 eV of KE V*eV= J The charge of He is -1 or 1.602*10^-19 C. 65000 eV = (-1)(1.602*10^-19C)(V) V= -4.05*10^23V
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19:13:27 65.0 keV is 65.0 * 10^3 eV, or 6.50 * 10^4 eV, of energy. The charge on a He nucleus is +2 e, where e is the charge on an electron. So assuming no dissipative forces, for every volt of potential difference, the He nuclues would gain 2 eV of kinetic energy. To gain 6.50 * 10^4 eV of energy the voltage difference would therefore be half of 6.50 * 10^4 voles, or 3.35 * 10^4 volts.
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RESPONSE --> The calculations made would have been correct if the correct nucleus charge was found. The charge needed is merely the number of electrons on the atom times the voltage or the potential difference inorder to get eV.
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19:20:21 Query gen phy text problem 17.18 potential 2.5 * 10^-15 m from proton; PE of two protons at this separation in a nucleus. What is the electrostatic potential at a distance of 2.5 * 10^-15 meters?
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RESPONSE --> U=PE PE=q1*q2*k / r = 1.6*10^-19*2*9*10^9 / 2.5*10^-15 PE=1153440 V
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19:25:52 STUDENT SOLUTION: For a part, to determine the electric potential a distance fo 2.5810^-15m away from a proton, I simply used the equation V = k q / r for electric potential for point charge: q = 1.60*10^-19C=charge on proton V = kq/r = 9.0*10^9N*m^2/C^2(1.60*10^-19C) / (2.5*10^-15m) = 5.8*10^5V. Part B was the more difficult portion of the problem. You have to consider a system that consists of two protons 2.5*10^-5m apart. The work done against the electric field to assemble these charges is W = qV. The potential energy is equal to the work done against the field. PE=(1.60*10^-19C)(5.8*10^5V) = 9.2*10^-14 J.
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RESPONSE --> I read the half problem wrong and thought that the PE was wanted between the two charges. which is in chapter 23 in the university book. Instead of considering two charges, only consider the test charge then use the formula.
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19:34:05 query univ phy 23.58 (24.58 10th edition). Geiger counter: long central wire 145 microns radius, hollow cylinder radius 1.8 cm. What potential difference between the wire in the cylinder will produce an electric field of 20,000 volts/m at 1.2 cm from the wire?
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RESPONSE --> The potential of the cylinder is V=R*E V= 200V for the cylinder The potential of the wire is v=2.9V The potential difference between the cylinder and the wire is200-2.9= 197.1V
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19:37:24 ** The voltage V_ab is obtained by integrating the electric field from the radius of the central wire to the outer radius. From this we determine that E = Vab / ln(b/a) * 1/r, where a is the inner radius and b the outer radius. If E = 20,000 V/m at r = 1.2 cm then Vab = E * r * ln(b/a) = 20,000 V/m * ln(1.8 cm / .0145 cm) * .012 m = 1157 V. **
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RESPONSE --> Since tha electrions move in a straight path the curve need to be integrated with equation that are pulled out from somewhere. Then the solution is easy to find by plugging in the values.
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19:41:00 Query univ 23.78 (24.72 10th edition). Rain drop radius .65 mm charge -1.2 pC. What is the potential at the surface of the rain drop?
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RESPONSE --> V= q k / r V= -1.66*10^23 V
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19:44:20 STUDENT RESPONSE FOLLOWED BY SOLUTION: The problem said that V was 0 at d = inifinity, which I understnad to mean that as we approach the raindrop from infinity, the potential differencegrows from 0, to some amount at the surface of the raindrop. Because water molecules are more positive on one side that the other, they tend to align in a certain direction. Since positive charges tend to drift toward negative charge, I would think that the raindrop, with its overall negative charge, has molecules arranged so that their more positive sides are pointing toward the center and negative sides will be alighed along the surface of the raindrop. Probably all wrong. I tried several differnet integrand configuraitons but never found one that gave me an answer in volts. SOLUTION: You will have charge Q = -1.2 * 10^-12 C on the surface of a sphere of radius .00065 m. The field is therefore E = k Q / r^2 = 9 * 10^9 N m^2 / C^2 * (-1.2 * 10^-12 C) / r^2 = -1.08 * 10^-2 N m^2 / C / r^2. Integrating the field from infinity to .00065 m we get (-1.08 * 10^-2 N m^2 / C) / (.00065 m) = -16.6 N m / C = -16.6 V. If two such drops merge they form a sphere with twice the volume and hence 2^(1/3) times the radius, and twice the charge. The surface potential is proportional to charge and inversely proportional to volume. So the surface potential will be 2 / 2^(1/3) = 2^(2/3) times as great as before. The surface potential is therefore 16.6 V * 2^(2/3) = -26.4 volts, approx.. **
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RESPONSE --> I used the wrong formula and should have been trying to find the electric field solving for E. E=k Q /r^2
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EwwȐߤ assignment #007 007. `Query 28 Physics II 06-22-2008
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20:48:07 Query introductory problems set 54 #'s 1-7. Explain how to obtain the magnetic field due to a given current through a small current segment, and how the position of a point with respect to the segment affects the result.
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RESPONSE --> To calculate a magnetic field. B= k' *IL / r^2 just like the original electric field equation except I stand for current and L stands for the length of the segment or the length the current travels. When points of reference are perpendicular to the segment the formula holds othermise the the equation must be multiplied by cons(theta). for the angle to the normal axis.
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20:48:39 ** 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). **
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RESPONSE --> The cos should be replaced by sin.
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20:50:31 Query 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?
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RESPONSE --> F=Q/V 7.00*10-9 * 12 = 8.4*10^-8 charges
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20:54:27 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.
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RESPONSE --> The capacitance is measured in F which is C/V. The equation however is C=Q/V. I think that the answer might be wrong because it says microF and not F. It microF is called 7*10^-9 F then I got the same answer.
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20:56:08 Explain how to obtain the magnetic field due to a circular loop at the center of the loop.
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RESPONSE --> The magnetic field is obtained in a similar way using B= k' *IL / r^2 I is the current flowing through the loop and L is the circumference of the loop. This works because the loop is always normal to the center.
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20:58:22 ** 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. **
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RESPONSE --> The circumference can also be incorporated into the formula using 2*pi *r
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21:00:27 Query magnetic fields produced by electric currents. What evidence do we have that electric currents produce magnetic fields?
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RESPONSE --> Electric currents produce magnetic fields because object in nature are somehow attracted to each other.
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21:01:39 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. **
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RESPONSE -->
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21:06:13 How is the direction of an electric current related to the direction of the magnetic field that results?
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RESPONSE --> The direction of the electric current and the magnetic field are always opposite because when cranking th generator and the current funs through the coils the current through the coils creates a magnetic field in the center with flux in the opposite direction of the current flow which causes resistance.
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21:07:03 Query problem 17.35 What would be the area of a .20 F capacitor if plates are separated by 2.2 mm of air?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. ok
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21:12:10 ** 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. **
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RESPONSE --> It just skipped something. When finding the electric field between two plates always use E= 4 pi k [q/a] to get the field then the voltage is obtained by multipling it by the distance between the plates. Now the capacitance can be found by c= q/v to get the area.
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21:15:32 Query 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.
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RESPONSE --> When the separation of the capacitor gets smaller thby half the capcitance doubles. so the new electric field would be stronger because it is closer together.
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21:17:30 ......!!!!!!!!...................................
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. ok i answered it right
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21:23:43 query 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?
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RESPONSE --> E= 4 pi k [q/a] E=8.48*10^10 V= E*d V=3.99*10^9V C=Q/V C=12/V= 3.01*10^-8F
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21:32:20 ** 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. **
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RESPONSE --> Now I'm confused. Why isnt any of this in the notes or experiments. I'am I suppose to teach this all to myself? The first part of the problem is completed correctly but for some reason the answers are completely different. This makes all of the other answers wrong. I thought the problem said the area was something else but I cannot remember. To calculate the average energy the max is divided by two then multiplied by the charge.
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21:40:00 query univ 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?
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RESPONSE --> E= 4 pi k [q/a] a=.0016 q=12v E=8.48*10^14 V= E*d V=7.97*10^12 C=Q/V C=12/V= 3.01*10^-8F The capcitance will be half the charge of the plate will be the same the field will be the same the energy store will be half.
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21:44:17 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.
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RESPONSE --> How is this happening? I followed the equation. the solution shows dividing 12 by .0047 instead of .0016. I thought it was suppose to be the area. The field is halved because it is double in distance like with the capacitance.If Capacitance is halved then so is Q.
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22:04:51 query univ 24.68 (25.52 10th edition). Solid conducting sphere radius R charge Q. What is the electric-field energy density at distance r < R from the center of the sphere? What is the electric-field energy density at distance r > R from the center of the sphere?
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RESPONSE --> Since electric fields are more powerful at the center the energy density or E/V will be greatest at the center as volume is decreases and E increases.When r
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22:09:15 ** The idea is that we have to integrate the energy density over all space. We'll do this by finding the total energy in a thin spherical shell of radius r and thickness `dr, using this result to obtain an expression we integrate from R to infinity, noting that the field of the conducting sphere is zero for r < R. Then we can integrate to find the work required to assemble the charge on the surface of the sphere and we'll find that the two results are equal. 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 Energy density = 1 / (8 pi k) E^2, Since your text uses epsilon0 I'll do the same on this problem, where the epsilon0 notation makes a good deal of sense: For the charged sphere we have for r > R 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 between r and r + `dr is nearly constant if `dr is small, with energy density 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 between distance r and r + `dr is therefore approximately 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 we approach 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 Q^2 / (8 pi epsilon0 R), which is the same as k Q^2 / (2 R). The work required to bring a charge `dq from infinity to a sphere containing charge q is k q / R `dq, leading to the integral of k q / R with respect to q. If we integrate from q = 0 to q = Q we get 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 k Q^2 / (2 R). Since k = 1 / (4 pi epsilon0) the work is k Q^2 / (2 R) = k Q^2 / (8 pi epsilon0 R). So the energy in the field is equal to the work required to assemble the charge distribution. **
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RESPONSE --> The sphere is a conducting sphere so there is no field within the field extends beyond its radius R. Beyond this point I have no further understanding. Where in the problem did it ask for anything involving work.
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υ银R܀ assignment #010 010. `Query 31 Physics II 06-22-2008
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22:24:40 Query Principles and General Physics 21.04. A circular loop of diameter 9.6 cm in a 1.10 T field perpendicular to the plane of the loop; loop is removed in .15 s. What is the induced EMF?
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RESPONSE --> Faradays E=dphi/ d time d phi is the change or instant magnetic flux = .096^2 * pi =.029 dtime= .15s the induced emf is .193V
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22:28:24 The average induced emf is the average rate of change of the magnetic flux with respect to clock time. The initial magnetic flux through this loop is flux = magnetic field * area = 1.10 T * (pi * .048 m)^2 = .00796 T m^2. The flux is reduced to 0 when the loop is removed from the field, so the change in flux has magnitude .0080 T m^2. The rate of change of
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RESPONSE --> The induced emf is the average rate of change in the flux over time. The initial flux is 1.10 * the area of the object. The flux over time is the average.
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22:28:41 flux is therefore .0080 T m^2 / (.15 sec) = .053 T m^2 / sec = .053 volts.
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RESPONSE -->
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22:36:04 query gen problem 21.23 320-loop square coil 21 cm on a side, .65 T mag field. How fast to produce peak 120-v output? How many cycles per second are required to produce a 120-volt output, and how did you get your result?
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RESPONSE --> flux = E* Area = .65* .21^2 =.028* 320 loops= 9.1728 V 120V divided by 9.17v is 13.08 seconds assuming 9.17 v/s The voltage can also be produced by having 13.08 cycles every sscond/
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22:39:12 The average magnitude of the output is peak output/sqrt(2) . We find the average output as ave rate of flux change. The area of a single coil is (21 cm)^2 = (.21 m)^2 and the magnetic field is .65 Tesla; there are 320 coils. When the plane of the coil is perpendicular to the field we get the maximum flux of fluxMax = .65 T * (.21 m)^2 * 320 = 19.2 T m^2. The flux will decrease to zero in 1/4 cycle. Letting t_cycle stand for the time of a complete cycle we have ave magnitude of field = magnitude of change in flux / change in t = 9.17T m^2 / (1/4 t_cycle) = 36.7 T m^2 / t_cycle. If peak output is 120 volts the ave voltage is 120 V / sqrt(2) so we have 36.7 T m^2 / t_cycle = 120 V / sqrt(2). We easily solve for t_cycle to obtain t_cycle = 36.7 T m^2 / (120 V / sqrt(2) ) = .432 second.+ A purely symbolic solution uses maximum flux = n * B * A average voltage = V_peak / sqrt(2), where V_peak is the peak voltage giving us ave rate of change of flux = average voltage so that n B * A / (1/4 t_cycle) = V_peak / sqrt(2), which we solve for t_cycle to get t_cycle = 4 n B A * sqrt(2) / V_peak = 4 * 320 * .65 T * (.21 m)^2 * sqrt(2) / (120 V) = .432 second.
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RESPONSE --> I obtained 9.17T m^2 not 19.2 I used the exact same formula and numbers. I'm confused where the 1/4 comes from. How do we know this?
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22:46:29 univ query 29.54 (30.36 10th edition) univ upward current I in wire, increasing at rate di/dt. Loop of height L, vert sides at dist a and b from wire. When the current is I what is the magnitude of B at distance r from the wire and what is the magnetic flux through a strip at this position having width `dr?
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RESPONSE --> B is = k' IL / r^2 the magnetic flux is simply the area L* dr * B
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22:50:32 ** The magnetic field due to the wire at distance r is 2 k ' I / r. The field is radial around the wire and so by the right-hand rule (thumb in direction of current, fingers point in direction of field) is downward into the page. The area of the strip is L * `dr. The magnetic flux thru the strip is therefore 2 k ' I / r * (L `dr). The total magnetic field over a series of such strips partitioning the area is thus sum(2 k ' I / r * L `dr, r from a to b). Taking the limit as `dr -> 0 we get } integral (2 k ' I / r * L with respect to r, r from a to b). Our antiderivative is 2 k ' I ln | r | * L; the definite integral therefore comes out to flux = 2 k ' L ln | b / a | * I. If I is changing then we have rate of change of flux = 2 k ' L ln | b / a | * dI/dt. This is the induced emf through a single turn. You can easily substitute a = 12.0 cm = .12 m, b = 36.0 cm = .36 m, L = 24.0 cm = .24 m and di/dt = 9.60 A / s, and multiply by the number of turns. **
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RESPONSE --> The magnetic field for the wire is not b it is 2k' l/r/ whre the magnetic field is perpendicular to the wire length-wise. The magnetic flux is then the field times the area it goes through.
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