course Phy 202 ??D???????v?x??????assignment #006
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19:04:19 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 --> This was question 17.2 135V - (-55V) = 190V
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19:05:45 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 didn't put the answer in J. I should have multiplied the Volts by the C of electrons. Also my sign was wrong I did not correctly note the direction.
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19:08:39 Query Principles and General Physics 17.8: Potential difference required to give He nucleus 65.0 keV of KE.
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RESPONSE --> 65 ke J/C = x J/ (2 * -1.6*10^-19C) x J = (65 J/C) (2* -1.6*10^-19C) x = -2.08 x 10^-17 J
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19:16:59 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 --> My answer was completely wrong and not even in the correct units. I tried to use the fromlual qVba instead of reasoning out the answer. 17.8 a) V = kQ/r V = ((9.0 * 10^9 Nm^2/C^2)(1.6 * 10^-19C))/(2.5*10^ -15m) V = 576000V b) V = kQ/r V = ((9.0 * 10^9 Nm^2/C^2)((2(1.6 * 10^-19C)))/ (2.5*10^-15m) V = 1152000 V
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19:16:59 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 -->
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19:16:59 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 -->
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19:18:39 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 did part A correctly but did part B in the same manner as A. I should have taken the voltage from A and multiplied it by the C of charge on electrongs to find the J of work.
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19:18:55 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 --> not assigned
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19:18:55 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 -->
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19:19:04 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 --> not assigned
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19:19:12 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 --> ok
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