course phy 202 10:3010/14/10 027.
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Given Solution: The change in potential is final potential - initial potential = -55 V - (135 V) = -190 V, so the change in the potential energy of the proton is -190 V * 1.6 * 10^-19 C = -190 J / C * 1.6 * 10^-19 C = -3.0 * 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 3.09 * 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 3.0 * 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qQuery Principles and General Physics 17.8: Potential difference required to give He nucleus 65.0 keV of KE. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 65 * 10^3 eV Positive 2 electrons gain 2 eV of KE. 6.50 * 10^4 eV of energy the voltage difference would therefore be half of 6.50 * 10^4 voles, or 3.35 * 10^4 volts confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&*2 ********************************************* Question: `qQuery gen phy text problem 17.18 potential 2.5 * 10^-15 m from proton; PE of two protons at this separation in a nucleus. What is the electrostatic potential at a distance of 2.5 * 10^-15 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: q = 1.60*10^-19C=charge proton V = kq/r = 9.0*10^9N*m^2/C^2(1.60*10^-19C) / (2.5*10^-15m) = 5.8*10^5 V 2 protons 2.5*10^-5m apart. PE=(1.60*10^-19C)(5.8*10^5V) 9.2*10^-14 Joules confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT QUESTION OK, got the first part. I follow the second, but it doesn't make much sense. So you multiplied your answer from the first by the charge again. INSTRUCTOR RESPONSE The given solution can benefit from an expanded explanation: Suppose you have two protons separated by a large, effectively infinite distance. They repel one another, so to move the protons closer to one another you would have to do positive work against the conservative electrostatic field. Just as when you lift an object against the conservative force of gravity, when you do work by 'pushing' an object against a conservative electrostatic field you increase its electrostatic PE. So you increase the PE of the system by 'pushing' one proton toward the other. The electrostatic potential at distance r from a point charge Q is V = k Q / r, and is defined to be the work per unit charge necessary to move another charge from infinite separation to that point. In the present case, the point charge Q is a proton and the electrostatic potential at a distance of 2.5*10^-15m from the proton is +5.8*10^5 volts, or +5.8 * 10^5 Joules / Coulomb. What this means is that to move charge from a large distance to this position requires +5.8 * 10^5 Joules for every Coulomb of charge you move. Now if the charge we are moving to that position is another proton, its charge is +1.6 * 10^-19 Coulombs. It therefore requires +5.8 Joules / Coulomb * 1.6 * 10^-19 Coulombs = 9.2 * 10^-14 Joules of work to get it there. This work is the potential energy of the system, relative to infinite separation. In symbols, V = k Q / r, and to move a second charge q to that point therefore requires work `dW = V * q = k Q q / r, and this is the electrostatic potential energy of the two-proton system. Going a little beyond this problem, here's a little 'enrichment': Now the two protons exert a lot of force on one another, trying to push one another away. At close distances they are held together by another force, not an electrostatic force. This force is called the 'strong' force. If the electrostatic force breaks free of the 'strong' force (as two protons would do almost instantly), the two protons will repel each other and the electrostatic force will quickly accelerate them to speeds close to that of light. From an energy point of view, the positive PE of the system will be converted to KE. Protons do manage to stay in this kind of close proximity in stable nuclei. Both protons and neutrons exert the 'strong' force on one another, if they are close enough. So for example if you get two protons and two neutrons into this sort of proximity, you get more 'strong' force and the electrostatic forces won't be able to break it. You end up with the very stable nucleus of Helium-4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&*1 ********************************************* Question: `qquery univ phy 23.58 (24.58 10th edition). Geiger counter: long central wire 145 microns radius, hollow cylinder radius 1.8 cm. What potential difference between the wire in the cylinder will produce an electric field of 20,000 volts/m at 1.2 cm from the wire? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** STUDENT QUESTION: Can you tell me what you integrated to get: E = Vab / ln(b/a) * 1/r ? INSTRUCTOR RESPONSE: Sure. The following assumes you know how to use Gaussian surfaces for axially symmetric charge distributions. If necessary see your text to fill in the details, but given the basic knowledge the explanation that follows is complete. I'll also be glad to clarify anything you wish to ask about: If the charge per unit length on the inner cylinder is lambda, then a coaxial cylinder of length L will contain charge Q = lamda * L. • So the flux through the cylinder will be 4 pi k Q = 4 pi k lambda * L. • Using symmetry arguments and assuming edge effects to be negligible, the electric field penetrates the curved surface of the cylinder at right angles. • The area of the curved surface of such a coaxial cylinder of radius r is 2 pi r * L, so the electric field is • field E = flux / area = 4 pi k lambda * L / (2 pi r L) = 2 k lambda / r. Integrating this field from inner radius a to outer radius b, we get the potential difference Vab: • Our antiderivative function is 2 k lambda ln | r |, so the change in the antiderivative is • Vab = 2 k lambda ( ln(b) - ln(a) ) = 2 k lambda ln(b / a). Thus Vab = 2 k lambda ln(b/a). • This gives us 2 k lambda = Vab / (ln(b/a)), which will be used below. • Since E = 2 k lambda / r, we substitute to get E = Vab / (ln(b/a)) * 1 / r, the expression about which you asked, and which we might wish to simplify into the form E = Vab / (r ln(b/a) ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qQuery univ 23.78 (24.72 10th edition). Rain drop radius .65 mm charge -1.2 pC. What is the potential at the surface of the rain drop? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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.. ** STUDENT QUESTION: I knew that my answer was off by some factor because the E decreased from when it was just one raindrop. I didn’t get that you would multiply it by two because the volume increased. ??? Can you explain why you would use the ratio of volume to radius increase in order to get the new E??? INSTRUCTOR RESPONSE: E depends on the total charge and the radius. When the two drops merge, their charges combine. This gives you double the charge compared to a single drop. We don't care about the volume, we care about the radius. However we know what happens to the volume: it doubles. So we use what we know about the volume to determine what happens to the radius: A sphere with twice the volume of another has 2^(1/3) times the radius. We end up with double the charge on a sphere with 2^(1/3) times the radius. Since the potential is proportional to the charge and inversely proportional to the radius, the potential changes by factor 2^(2/3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*