course Phy 202
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09:33:26 Note that my solutions use k notation rather than epsilon0 notation. The k notation is more closely related to the geometry of Gauss' Law and is consistent with the notation used in the Introductory Problem Sets. However you can easily adapt the two notations, since k = 1 / (4 pi epsilon0), or alternatively epsilon0 = 1 / (4 pi k).
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RESPONSE --> ok
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09:35:26 Query RC circuits. Explain why the current in a discharging RC circuit, consisting of an initially charged capacitor discharging through a resistor, decreases exponentially with time.
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RESPONSE --> The current is the rate at which charge changes. The exponential decrease results because the current is proportional to the voltage which is proportional to the charge.
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09:37:48 ** The exponential decrease results because the current is proportional to the voltage, which is proportional to the charge. The current is the rate at which charge changes. Thus the rate at which the charge changes is proportional to the charge. Whenever the rate at which something changes is proportional to that thing (e.g., population of a species in a noncompeting environment, money in an interest-bearing account) the change will be exponential. **
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RESPONSE --> ok i understand
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09:41:53 Explain why the current in a charging RC circuit, consisting of an initially uncharged capacitor in parallel with a resistor, decreases exponentially with time.
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RESPONSE --> The current is changing because the current flow increases the charge on the capacitor. So the more the charge the more opposition from the voltage and less current flow.
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09:43:07 STUDENT RESPONSE WITH INSTRUCTOR COMMENT: This may not be right. Electrons flow out form the negative terminal of the battery in the charging RC circuit, through the resistor and accumulate on the upper plate of the capacitor. The electrons will flow into the positive terminal which leaves a positive charge on the other plate of the capacitor. This charge accumulates on the capacitor the potential difference across it increases and the current decreases until the capacitor equals the emf of the battery. INSTRUCTOR COMMENT: This is close to a complete answer. The current flow increases the charge on the capacitor, which results in a voltage that opposes the source; the more charge the more opposition and the less current flow. **
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RESPONSE --> ok i understand
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09:47:05 Query gen phy 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 --> Vkq/r=9.0*10^9N*m^2/C62(1.60*10^-19C)/2.5*10^-15m = 5.9*10^5V. The work done by the electric field is W = qV. PE=(1.60*10^-19C)(5.8*10^5V)=9.24*10^-14J.
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09:49:18 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/C62(1.60*10^-19C)/2.5*10^-15m = 5.8*10^5V. Part B was the more difficult potrion of the problem. You have to consider a system that consists of two protons 2.5*10^-5m apart. The work done by the electric field is W = qV. Thus the potential energy will be equal to the work done by the system. PE=(1.60*10^-19C)(5.8*10^5V) =9.2*10^-14J.
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RESPONSE --> ok i understand
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09:51:21 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 --> dna
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09:51:24 ** The voltage would be 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 --> dna
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09:51:27 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 --> dna
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09:51:30 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 --> dna
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