#$&* course Phy 122 9 pm July 15 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: ** IL is the source. The law is basically an inverse square law and the angle theta between IL and the vector r from the source to the point also has an effect so that the field is B = k ' I L / r^2 * sin(theta). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qQuery principles and general college physics problem 17.34: How much charge flows from each terminal of 7.00 microF capacitor when connected to 12.0 volt battery? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Q = C * V = 7 micro F * 12 v = 7 micro C / volt * 12 v = 84 micro C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 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: ok ********************************************* Question: Explain how to obtain the magnetic field due to a circular loop at the center of the loop. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: B = k I / r ^2 sum (dL) B = 2 pi r k I / r ^2 = 2 pi k I / r confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For current running in a circular loop: Each small increment `dL of the loop is a source I `dL. The vector from `dL to the center of the loop has magnitude r, where r is the radius of the loop, and is perpendicular to the loop so the contribution of increment * `dL to the field is k ' I `dL / r^2 sin(90 deg) = I `dL / r^2, where r is the radius of the loop. The field is either upward or downward by the right-hand rule, depending on whether the current runs counterclockwise or clockwise, respectively. The field has this direction regardless of where the increment is located. The sum of the fields from all the increments therefore has magnitude B = sum(k ' I `dL / r^2), where the summation occurs around the entire loop. I and r are constants so the sum is B = k ' I / r^2 sum(`dL). The sum of all the length increments around the loop is the circumference 2 pi r of the loop so we have B = 2 pi r k ' I / r^2 = 2 pi k ' I / r. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qWhat would be the area of a .20 F capacitor if plates are separated by 2.2 mm of air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: A = 4 pi k d * C = 4 pi * 9 * 10 ^9 / C ^2 * .20 C / V * .0022 m = = 5 * 10 ^7 / C ^2 * C / ( J / C) * m = 5 * 10 ^7 / (N m) * m = 5 * 10 ^7 m ^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!