course phy 202
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16:07:38 Query problem 16.32. field 745 N/C midway between two equal and opposite point charges separated by 16 cm. What is the magnitude of each charge?
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RESPONSE --> I know you would use E= kQ/r^2 i think you would just plug in the information: 745 N/C = ( mag * (50 * 10^-6 C))/ (.16 m)^2 =3.8 * 10^5
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16:10:35 ** The charges are each 8 cm from the field point. If the magnitude of the charge is q then the field contribution of each charge is k q / r^2, with r = 8 cm = .08 meters. Since both charges contribute equally to the field, with the fields produced by both charges being in the same direction (on any test charge at the midpoint one force is of repulsion and the other of attraction, and the charges are on opposite sides of the midpoint), the field of either charge has magnitude 1/2 (745 N/C) = 372.5 N/C. Thus E = 372.5 N/C and E = k q / r^2. We know k, E and r so we solve for q to obtain q = E * r^2 / k = 372.5 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2) = 372.5 N/C * .0064 m^2 / (9 * 10^9 N m^2 / C^2) = 2.6 * 10^-10 C, approx. **
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RESPONSE --> I just about had it but i did not multiply 745 N/C by .5 for the field of either charge and i solved for the wrong thing
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16:11:03 If the charges are represented by Q and -Q, what is the electric field at the midpoint?
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RESPONSE --> It would be 2 k Q / r^2
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16:11:25 ** this calls for a symbolic expression in terms of the symbol Q. The field would be 2 k Q / r^2, where r=.08 meters and the factor 2 is because there are two charges of magnitude Q both at the same distance from the point. **
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RESPONSE --> Ok this is what i thought
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16:12:38 Query electrostatics. Explain why a charge enclosed in an aluminum cylinder is not affected by a charge outside the aluminum cylinder, when the charges would clearly affect one another in the absence of the cylinder.
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RESPONSE --> I did not have the PVC pipe so i am waiting to see what else i could use. I don't want to have to go buy a large piece of PVC pipe if i only need a bit. But from what i understand the aluminum
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16:13:20 STUDENT RESPONSE AND INSTRUCTOR COMMENT: The electrons in a conductor will freely move and redidtribute them selves so that the field can not penetrate the cylinder. INSTRUCTOR COMMENT: Good explanation. See the next paragraph for a more detailed explanation: the aluminum cylinder is full of free charges that can migrate freely anywhere on the cylinder. The outside charge attracts opposite charges, which build to greatest density on regions of the aluminum cylinder nearest it. This process must continue until there is no electric field in the aluminum, because if there is an electric field in the aluminum then charges will move in response to it. Movement of charges must continue until the field is eliminated. This effectively 'shields' all point inside the cylinder from the effect of external charges.
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RESPONSE --> OK
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16:13:52 Explain why bringing a charged plastic rod near the far end of an aluminum rod will have a direct affect on a charge close to the near end of the rod, whereas the same charge and plastic rod will exhibit no measurable interaction in the absence of the aluminum rod.
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RESPONSE --> I am not sure why this happens but i will read the solution and will take note of what happens
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16:17:32 ** This question concerns the situation where a charged object is located near one end of the metal rod and the charged plastic rod is brought near the other end of the metal rod. Nothing ever touches anything else so there is no migration of charge from one object to the other. When the charged plastic rod is brought near there is however a redistribution of charge on the conducting rod and the force on the charged object will change. The redistribution of charges causes the end of the pipe near the rod to take on a charge opposite to that of the rod; this occurs by displacement of charge from the other end of the rod, which therefore ends up with the same type of charge as the plastic rod. Thus the charge near that end will experience a force in the same direction as if the rod was near. If the metal rod wasn't there, the distance of the plastic rod would cause the effect on that charge to be minimal. **
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RESPONSE --> Ok so when the charged plastic rod is brought near the metal rod is however a redistribution of charge on the conducting rod and the force on the charged object will change and the redistribution of charges causes the end of the pipe near the rod to take on a charge opposite to that of the rod. This occurs by displacement of charge from the other end of the rod, which therefore ends up with the same type of charge as the plastic rod.
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16:17:36 query univ 22.30 (10th edition 23.30). Cube with faces S1 in xz plane, S2 top, S3 right side, S4 bottom, S5 front, S6 back on yz plane. E = -5 N/(C m) x i + 3 N/(C m) z k. What is the flux through each face of the cube, and what is the total charge enclosed by the cube?
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16:17:37 ** Flux is not a vector quantity so your flux values will not be multiples of the i, j and k vectors. The normal vectors to S1, S2, ..., S6 are respectively -j, k, j, -k, i and -i. For any face the flux is the dot product of the field with the normal vector, multiplied by the area. The area of each face is (.3 m)^2 = .09 m^2) So we have: For S1 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-j) * .09 m^2 = 0. For S2 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( k) * .09 m^2 = 3 z N / (C m) * .09 m^2. For S3 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( j) * .09 m^2 = 0. For S4 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-k) * .09 m^2 = -3 z N / (C m) * .09 m^2. For S5 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( i) * .09 m^2 = -5 x N / (C m) * .09 m^2. For S6 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-i) * .09 m^2 = 5 x N / (C m) * .09 m^2. On S2 and S4 we have z = .3 m and z = 0 m, respectively, giving us flux .027 N m^2 / C on S2 and flux 0 on S4. On S5 and S6 we have x = .3 m and x = 0 m, respectively, giving us flux -.045 N m^2 / C on S5 and flux 0 on S6. The total flux is therefore .027 N m^2 / C - .045 N m^2 / C = -.018 N m^2 / C. Since the total flux is 4 pi k Q, where Q is the charge enclosed by the surface, we have 4 pi k Q = -.018 N m^2 / C and Q = -.018 N m^2 / C / (4 pi k) = -.018 N m^2 / C / (4 pi * 9 * 10^9 N m^2 / C^2) = -1.6 * 10^-13 C, approx. **
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16:17:39
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16:17:42 query univ 22.37 (23.27 10th edition) Spherical conducting shell inner radius a outer b, concentric with larger conducting shell inner radius c outer d. Total charges +2q, +4q. Give your solution.
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16:17:45 ** The electric field inside either shell must be zero, so the charge enclosed by any sphere concentric with the shells and lying within either shell must be zero, and the field is zero for a < r < b and for c < r < d. Thus the total charge on the inner surface of the innermost shell is zero, since this shell encloses no charge. The entire charge 2q of the innermost shell in concentrated on its outer surface. For any r such that b < r < c the charge enclosed by the corresponding sphere is the 2 q of the innermost shell, so that the electric field is 4 pi k * 2q / r^2 = 8 pi k q / r^2. Considering a sphere which encloses the inner but not the outer surface of the second shell we see that this sphere must contain the charge 2q of the innermost shell. Since this sphere is within the conducting material the electric field on this sphere is zero and the net flux thru this sphere is zero. Thus the total charge enclosed by this sphere is zero. Since the charge enclosed by the sphere includes the 2q of the innermost shell, the sphere must also enclose a charge -2 q, which by symmetry must be evenly distributed on the inner surface of the second shell. Any sphere which encloses both shells must enclose the total charge of both shells, which is 6 q. Since we have 2q on the innermost shell and -2q on the inner surface of the second shell the charge on the outer surface of this shell must be 6 q. For any r such that d < r the charge enclosed by the corresponding sphere is the 6 q of the two shells, so that the electric field is 4 pi k * 6q / r^2 = 24 pi k q / r^2. **
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16:17:47 query univ 23.46 (23.34 10th edition). Long conducting tube inner radius a, outer b. Lin chg density `alpha. Line of charge, same density along axis.
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16:17:49 ** Gaussian surfaces for this configuration are cylinders of radius r and 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. For r < a the charge enclosed by the Gaussian surface is L * alpha so that the flux is 4 pi k L * alpha and the electric field is 4 pi k L * alpha / (2 pi r L ) = 2 k alpha / r. For a < r < b the Gaussian surface lies within the conductor so the field is zero. This implies that 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 charge -L * alpha, so that the inner surface is characterized by 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 density includes that of the line charge as well as the inner surface of the shell the outer surface of the shell has charge density 2 alpha. **
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SՈրԌ assignment #027 ]ĕ圯N Physics II 11-09-2005
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18:53:14 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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18:53:40 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 --> Because the current is proportional to the voltage
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18:54:03 ** 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 --> Exactly, the rate at which the charge changes is proportional to the charge
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18:54:16 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 --> because the flow increases the charge in the capacitor and therefore the voltgae will go against the source used
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18:54:38 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 think i understand this concept
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18:57:03 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 --> You first will use V = kq/r = 9.0 * 10^9N*m^2/62(1.60 * 10^-19)/2.5*10^-15 = 5.8*10^5
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18:58:08 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 forgot to do the part B and it was really hard to do, but my understanding of this is that since 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
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18:58:11 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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18:58:13 ** 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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18:58:15 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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18:58:17 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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