course Phy 202 I don't have the capacitor/generator kit, so I thought I would go through and do all of the practice problems and other work I could get done for this week, and if you approve an extension the first thing I will do is order the set and complete the labs. If you think I should get the capacitor and generator before you grant the extension, please advise. Thanks! S]Յ֧]qRͥd݊assignment #006
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22:26:45 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 --> W_ba = q(V_b - V_a) W_ba = 1.602*10^-19 C (-55 V - 135 V) W_ba = 3.04*10^-17 J
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22:28:17 07-30-2007 22:28:17 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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NOTES -------> For some reason I used 135 instead of 125 in my work, but I think I got the general idea correct. I'm going to save this as notes.
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22:39:43 Query Principles and General Physics 17.8: Potential difference required to give He nucleus 65.0 keV of KE.
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RESPONSE --> q = 2e = 3.204*10^-19 C 65 keV = abs value(65000 V * -1.602*10^-19 C) = 1.0413*10^-14 J = W_ba V_ba = 1.0413*10^-14 J/3.2*10^-19 C = 3.25*10^4
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22:44:49 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 --> My methodology was totally off on this but I ended up with the correct answer. 6.5*10^4 eV q = 2e For every V, He nucleus gains 2 eV of K.E. To gain 6.50 * 10^4 eV of energy, the difference would be the inverse of the KE/V, or 1/2 of 6.50*10^4 V. The way I did it makes more sense to me. Was it a fluke that my answer was also correct, or could I use this methodology to get the same result always?
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23:09:04 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 --> r = 2.5*10^-15 m q = 1.602*10^-19 C V = k(Q/r) = 9.0*10^9 N*m^2/C^2*(1.602*10^-19 C / 2.5*10^-15 m) V = 5.77*10^5 V PE = k*(Q_1*Q_2)/r PE = 9.0*10^9 N*m^2/C^2 * (1.602*10^-19)^2 C^2 / 2.5*10^-15 m = 9.24*10^-14 J V = PE/q V = 9.25*10^-14 J/1.602*10^-19 C = 1.48*10^-32 V ??
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23:09:57 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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23:10:02 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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23:10:06 ** 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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23:10:08 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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23:10:20 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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