course PHY 202 ???????Q??r???assignment #007
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15:05:15 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 --> thw work done on the charge would equal the electric potential multiplied by the charge of the feild.
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15:13:16 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 --> I was correct in finding the eletric potential howver I forgot that I was working with a proton in which the charge is given. had I known that I would have multiplied the two to get my answer.
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???pz|??????????assignment #007 007. `Query 28 Physics II 06-18-2008
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15:28:34 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 --> The Magnetic feild ""B"" is found by multiplying k by the source of the magnetic feild or IL and then dividing by the distance of the source of the feild to the point in question. or B= k IL/ r^2. The position of the point is important because the whole feild is dependent not only on the distance of the points but of the orientation of the charge flow with respect to the feild.
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15:29:54 ** 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 --> I missed the sin theta bit, but was right about everything else. It makes sense to need to include sin theta beccause of the question of orientation of the point to the magnetic feild playing such a key role.
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15:35:19 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 --> to find this the equation Q=CV is needed. V is equal to 12 volts and C or the capacitance has been described as 7.00 microF or 7.00 x 10^ -6. This being the case the answer is 8.4 x 10^ -5.
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15:36:12 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 --> I converted the mirco F out when I didn't need to, resulting in way too many 0s.
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15:40:15 Explain how to obtain the magnetic field due to a circular loop at the center of the loop.
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RESPONSE --> B= F/IL.
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15:43:02 ** 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 --> I was nowhere near the correct answer. But by using the geometry of a circle and the summation of all the vectors of a small sample of the circle it is possible to find r. I have the equation that is needed to get this answer but I don't understand where it came from.
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15:44:38 Query magnetic fields produced by electric currents. What evidence do we have that electric currents produce magnetic fields?
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RESPONSE --> When a compass needle is brought near a wire that carries as electric current the needle deflects, because a compass needle is deflected by a magnetic feild this proves tha electric currents do have some connection to magnetism.
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15:45:13 How is the direction of an electric current related to the direction of the magnetic field that results?
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RESPONSE --> got it.
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15:45:31 ** 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. **
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RESPONSE --> got that too.
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16:00:06 06-18-2008 16:00:06 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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NOTES -------> to solve this the equation C = eo(A/D) should be rewritten to solve for Z. From this wek get Cd/E0. Plugging .20F into the capacitance and .0022 in for the d ( e is a constant of 8.85 x 10^ -12) the answer is 4.97 x 10 ^7
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16:00:07 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 -->
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16:01:01 ** 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 --> I was entirely wrong but I'm not sure why you use this equation instead of the one that I worked out of the book.
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17:48:04 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 --> Because the seperation is halved, the charge Q will remain the same, however, the capacitance increases as a result of the dielectric. Because there is no change in the charge and C in icreased, to conserve energy in the equation Q =CV the voltage must decrease.
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17:48:42 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. ** 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, ration 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. **
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RESPONSE --> I was in the correct line of thinking.
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17:50:53 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 --> The capacitance is found by the equation C = e (A/D) e is a constant, A is converted to meters to equal .16 and the distance is also converted to .oo47. From this the equation is solved to where C = 3.06 x 10^-10
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17:52:04 ** 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 --> again I seem to have a different equationthat I've gotten from the book and it came up with a different answer. Ido however understand the logic behind the answer given.
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