#$&* course Phy 232 7/28 12 027.
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Given Solution: The change in potential is final potential - initial potential = -55 V - (135 V) = -190 V, so the change in the potential energy of the proton is -190 V * 1.6 * 10^-19 C = -190 J / C * 1.6 * 10^-19 C = -3.0 * 10^-17 J. In the absence of dissipative forces this is equal and opposite to the change in the KE of the proton; i.e., the proton would gain 3.09 * 10^-17 J of kinetic energy. Change in potential energy is equal and opposite to the work done by the field on the charge, so the field does 3.0 * 10^-17 J of work on the charge. Since the charge of the proton is equal in magnitude to that of an electron, he work in electron volts would be 180 volts * charge of 1 electron= 180 eV. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qQuery Principles and General Physics 17.8: Potential difference required to give He nucleus 65.0 keV of KE. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 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. STUDENT QUESTION I didn’t convert the keV into eV. What do these units even mean?? INSTRUCTOR RESPONSE k means 'kilo'; so a keV is 10^3 eV. An electron volt is the PE change of an electron as it moves through a PE change of +1 volt. A Joule is the PE change of a Coulomb of negative charge as it moves through a PE change of +1 volt. Since the charge of an electron has magnitude 1.6 * 10^-19 Coulomb, an electron volt is 1.6 * 10^-19 Joules. Within the electron shell of an atom, and in many other applications, we are dealing with charges of in small whole-number multiples of 1.6 * 10^-19 Coulomb, moving between points where potential changes are anywhere from a few millivolts to a few volts. The associated energy changes are more easily thought of in terms of electron volts (typical values might range from .001 eV to around 100 eV) than ergs or Joules (where the same range might be, for example, 10^-17 Joules to 10^-22 Joules). Numbers between .001 and 100 are easy to think about, an relate easily to the energy of a single electron moving through a potential difference of a single volt. Numbers like 10^-17 and 10^-22 are harder to imagine. STUDENT QUESTION I understand the formula, but didn’t know where to find He? May have missed it in the notes……anyway INSTRUCTOR RESPONSE It is assumed to be general knowledge that a helium nucleus contains 2 protons, each with a positive charge equal in magnitude to the electron charge. However not everyone remembers this, and without this knowledge it would be difficult to get the entire solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* 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: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT SOLUTION: For a part, to determine the electric potential a distance of 2.58 * 10^-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. WORTH KNOWING BUT NOT NECESSARY AT THIS POINT: 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. STUDENT QUESTION: A little confused about the V=k*Q/r, where it looks like I can change the Q to q and substitute? INSTRUCTOR RESPONSE There is nothing special about Q and q. The charge can be called Q or q or q1 or q2, or anything else. The statement 'The electrostatic potential at distance r from a point charge Q is V = k Q / r' uses Q for the charge. Had the statement been 'The electrostatic potential at distance r from a point charge q1 is V = k q1 / r' it would have been equally valid. Either definition tells you that whatever the number of symbol you choose to use for the charge, that symbol goes between k and the /. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qquery univ 22.38 11th edition 23.46 (23.34 10th edition). Long conducting tube inner radius a, outer b. Lin chg density `alpha (or possibly `lambda, depending on which edition of the text you are using). Line of charge, same density along axis. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: (a) (i) For r < a, Gauss’s law gives E(2*pi*r*l)=Qencl/epsilon0 =alpha*l/epsilon0. And E = alpha/(2*pi*epsilon0*r) (ii) The electric field is zero because these points are within the conducting material. (iii) For r >b , Gauss’s law gives E(2*pi*r*l) = Qencl/epsilon0 =2*alpha*l/epsilon0 And E = alpha/(pi*epsilon0*r) (b) (i) The Gaussian cylinder with radius r, for a
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Given Solution: The Gaussian surfaces appropriate to this configuration are cylinders of length L which are coaxial with the line charge. The symmetries of the situation dictate that the electric field is everywhere radial and hence that the field passes through the curved surface of each cylinder at right angle to that surface. The surface area of the curved portion of any such surface is 2 pi r L, where r is the radius of the cylinder. For r < a the charge enclosed by the Gaussian surface is L * alpha so that the flux is charge enclosed = 4 pi k L * alpha and the electric field is electric field = flux / area = 4 pi k L * alpha / (2 pi r L ) = 2 k alpha / r. For a < r < b, a Gaussian surface of radius r lies within the conductor so the field is zero (recall that if the field wasn't zero, the free charges inside the conductor would move and we wouldn't be in a steady state). So the net charge enclosed by this surface is zero. Since the line charge enclosed by the surface is L * alpha, the inner surface of the conductor must therefore contain the equal and opposite charge -L * alpha, so that the inner surface carries charge density -alpha. For b < r the Gaussian surface encloses both the line charge and the charge of the cylindrical shell, each of which has charge density alpha, so the charge enclosed is 2 L * alpha and the electric field is radial with magnitude 4 pi k * 2 L * alpha / (2 pi r L ) = 4 k alpha / r. Since the enclosed charge that of the line charge (L * alpha) as well as the inner surface of the shell (L * (-alpha) ), which the entire system carries charge L * alpha, we have line charge + charge on inner cylinder + charge on outer cylinder = alpha * L, we have alpha * L - alpha * L + charge on outer cylinder = alpha * L, so charge on outer cylinder = 2 alpha * L, so the outer surface of the shell has charge density 2 alpha. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `qquery univ phy 23.62 11th edition 23.58 (24.58 10th edition). Geiger counter: long central wire 145 microns radius, hollow cylinder radius 1.8 cm. What potential difference between the wire in the cylinder will produce an electric field of 20,000 volts/m at 1.2 cm from the wire? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: E = Vab/[ln(b/a)*r] A = 145*10^-6; b = 0.018 Vab = (2*10^4)*(ln(.018/(145*10^-6))*.012 = 1157 V confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `qQuery univ 23.80 11th edition 23.78 (24.72 10th edition). Rain drop radius .65 mm charge -1.2 pC. What is the potential at the surface of the rain drop? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: V = k*Q/R = k*(-1.2*10^-12)/(6.5*10^-4) = -16.6 V confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique rating #$&* ok "