#$&* course Phy 7/8 5 Query 24024.
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Given Solution: `aSTUDENT RESPONSE AND INSTRUCTOR COMMENT: electric field is the concentration of the electrical force ** That's a pretty good way to put it. Very specifically electric field measures the magnitude and direction of the electrical force, per unit of charge, that would be experienced by a charge placed at a given point. ** STUDENT COMMENT: Faraday explain that it reached out from the charge, so would that be a concentration? It seems to me that the concentration would be near the center of the charge and the field around it would be more like radiation extending outward weakening with distance. INSTRUCTOR RESPONSE That's a good, and very important, intuitive conception of nature of the electric field around a point charge. However the meaning of the field is the force per unit charge. If you know the magnitude and direction of the field and the charge, you can find the magnitude and direction of the force on that charge. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qExplain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane due to a given point charge at the origin. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F = k q1 * Q / r^2 F / Q = k q1 / r^2 sign of q1 (positive or negative) indicates the direction confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The electric field is therefore F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge. The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative). The direction of the electric field is in the direction of the displacement vector from the origin to the point if q1 is positive, and opposite to this direction if q1 is negative. To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If q1 is positive then this is the direction of the field. If q1 is negative then the direction of the field is opposite this direction, 180 degrees more or less than the calculated angle. ** STUDENT QUESTION Why is it just Q and not Q2? INSTRUCTOR RESPONSE q1 is a charge that's actually present. Q is a 'test charge' that really isn't there. We calculate the effect q1 has on this point by calculating what the force would be if a charge Q was placed at the point in question. This situation can and will be expanded to a number of actual charges, e.g., q1, q2, ..., qn, at specific points. If we want to find the field at some point, we imagine a 'test charge' Q at that point and figure out the force exerted on it by all the actual charges q1, q2, ..., qn. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Principles of Physics and General Physics problem 16.15 charges 6 microC on diagonal corners, -6 microC on other diagonal corners of 1 m square; force on each. What is the magnitude and direction of the force on the positive charge at the lower left-hand corner? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F = 9 * 10^9 N m^2/C^2 (6 * 10^-6 C)(6 * 10^-6 C) / ( 1 m)^2 = 324 * 10^-3 N F = 9 * 10^9 N m^2/C^2( 6 * 10^-6 C)( 6* 10^-6 C) / ( 1.414 m)^2 = 162 * 10^-3 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The charges which lie 1 meter apart are unlike and therefore exert attractive forces; these forces are each .324 Newtons. This is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1 m)^2 = 324 * 10^-3 N = .324 N. Charges across a diagonal are like and separated by `sqrt(2) meters = 1.414 meters, approx, and exert repulsive forces of .162 Newtons. This repulsive force is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6* 10^-6 C) / ( 1.414 m)^2 = 162 * 10^-3 N = .162 N. The charge at the lower left-hand corner therefore experiences a force of .324 Newtons to the right, a force of .324 Newtons straight upward and a force of .162 Newtons at 45 deg down and to the left (at angle 225 deg with respect to the standard positive x axis, which we take as directed toward the right). This latter force has components Fy = .162 N sin(225 deg) = -.115 N, approx, and Fx = .162 N cos(225 deg) = -.115 N. The total force in the x direction is therefore -.115 N + .324 N = .21 N, approx; the total force in the y direction is -.115 N + .324 N = .21 N, approx. Thus the net force has magnitude `sqrt( (.21 N)^2 + (.21 N)^2) = .29 N at an angle of tan^-1( .21 N / .21 N) = tan^-1(1) = 45 deg. The magnitude and direction of the force on the negative charge at the lower right-hand corner is obtained by a similar analysis, which would show that this charge experiences forces of .324 N to the left, .324 N straight up, and .162 N down and to the right. The net force is found by standard vector methods to be about .29 N up and to the left. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qquery university physics 21.66 / 21.72 11th edition 21.68 (22.52 10th edition) 5 nC at the origin, -2 nC at (4 cm, 0). If 6 nC are placed at (4cm, 3cm), what are the components of the resulting force? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -2nC charge@3 cm above the 6 nC will have: f = 9 * 10^9 N m^2 / C^2(-2 * 10^-9 C) ( 6 * 10^-9 C) / (.03 m)^2 = .00012 N. vector = -.00012 N j_hat. 5 nC charge@4 cm from the 6 nC charge will have: f = 9 * 10^9 N m^2 / C^2(6 * 10^-9 C) ( 5 * 10^-9 C) / (.05 m)^2 = .000108 N direction: 4i+3j unit vector = 4/5i + 3/5j 000108 N(4/5i + 3/5j) = .000086 Ni + .000065 Nj. The resultant force = .000086 Ni - .000055 Nj. Magnitude = sqrt( (.000086 N)^2 + (-.000055 N)^2 ) = .00011 N, Angle = arctan(-.000055 N / (.000086 N) ) = -33 degrees confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The -2 nC charge lies 3 cm above the 6 nC charge, so it exerts force 9 * 10^9 N m^2 / C^2 * (-2 * 10^-9 C) ( 6 * 10^-9 C) / (.03 m)^2 = .00012 N. The force between the two charges is a force of attraction, so the direction of the force on the 6 nC charge is the negative y direction. The vector force is thus -.00012 N * `j. The 5 nC charge lies at distance 4 cm from the 6 nC charge, so it exerts force 9 * 10^9 N m^2 / C^2 * (6 * 10^-9 C) ( 5 * 10^-9 C) / (.05 m)^2 = .000108 N, approx... The charges repel, so this force acts in the direction of the vector 4 `i + 3 `j representing the displacement from the origin to the point (4 cm, 3 cm). The unit vector in this direction is easily seen to be 4/5 `i + 3/5 `j = .8 `i + .6 `j. It follows that the force vector is .000108 N ( 4/5 `i + 3/5 `j) = .000086 N * `i + .000065 N * `j. The resultant force is therefore the sum of these two vectors, which is about .000086 N * `i - .000055 N * `j. This vector has magnitude sqrt( (.000086 N)^2 + (-.000055 N)^2 ) = .00011 N, approx., and angle arcTan(-.000055 N / (.000086 N) ) = -33 degrees, approx., or 360 deg - 33 deg = 327 deg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery univ phy 21.78 / 21.80 11th edition 21.76 (10th edition 22.60) quadrupole (q at (0,a), (0, -a), -2q at origin). For y > a what is the magnitude and direction of the electric field at (0, y)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: kq/(y - a)^2 + kq/(y + a)^2 - 2kq/y^2 = 2kq(y^2 + a^2)/((y + a)^2(y - a)^2) - 2kq/y^2 = 2kq[(y^2 + a^2)y^2 - (y+a)^2 ( y-a)^2) ] / ( y^2 (y + a)^2(y - a)^2) = 2kq[y^4 + a^2 y^2 - (y^2 - a^2)^2 ] / ( y^2 (y + a)^2(y - a)^2) = 2kq[y^4 + a^2 y^2 - y^4 + 2 y^2 a^2 - a^4 ] / ( y^2 (y + a)^2(y - a)^2) = 2kq[ 3 a^2 y^2 - a^4 ] / ( y^2 (y + a)^2(y - a)^2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The magnitude of the field due to the charge at a point is k q / r^2. For a point at coordinate y on the y axis, for y > a, we have distances r = y-a, y+a and y. The charges at these distances are respectively q, q and -2q. So the field is k*q/(y - a)^2 + k*q/(y + a)^2 - 2k*q/y^2 = 2*k*q*(y^2 + a^2)/((y + a)^2*(y - a)^2) - 2*k*q/y^2 = 2*k*q* [(y^2 + a^2)* y^2 - (y+a)^2 ( y-a)^2) ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - (y^2 - a^2)^2 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - y^4 + 2 y^2 a^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [ 3 a^2 y^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) . For large y the denominator is close to y^6 and the a^4 in the numerator is insignifant compared to a^2 y^2 sothe expression becomes 6 k q a^2 / y^4, which is inversely proportional to y^4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery univ 21.104 / 22.102 annulus in yz plane inner radius R1 outer R2, charge density `sigma. What is a total electric charge on the annulus? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Q = sigma * A = sigma * (pi R2^2 - pi R1^2). dQ = 2pir *dr * sigma From any small segment of this ring the electric field at a point of the x axis would be directed at angle arctan( r / x ) with respect to the x axis. By either formal trigonometry or similar triangles we see that the component parallel to the x axis will be in the proportion x / sqrt(x^2 + r^2) to the magnitude of the field from this small segment. By symmetry only the xcomponent of the field will remain when we sum over the entire ring. field due to the ring: k `dQ / (x^2 + r^2) * x / sqrt (x^2 + r^2) = 2pikr `dr * x / (x^2 + r^2)^(3/2). magnitude of field = integral from R1 to R2 ( 2 pi k r x /(x^2 + r^2)^(3/2) with respect to r magnitude of field = 2pikx* | 1 /sqrt(x^2 + r1^2) - 1 / sqrt(x^2 + r2^2) | confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The total charge on the annulus is the product Q = sigma * A = sigma * (pi R2^2 - pi R1^2). To find the field at distance x along the x axis, due to the charge in the annulus, we first find the field due to a thin ring of charge: The charge in a thin ring of radius r and ring thickness `dr is the product `dQ = 2 pi r `dr * sigma of ring area and area density. From any small segment of this ring the electric field at a point of the x axis would be directed at angle arctan( r / x ) with respect to the x axis. By either formal trigonometry or similar triangles we see that the component parallel to the x axis will be in the proportion x / sqrt(x^2 + r^2) to the magnitude of the field from this small segment. By symmetry only the xcomponent of the field will remain when we sum over the entire ring. So the field due to the ring will be in the same proportion to the expression k `dQ / (x^2 + r^2). Thus the field due to this thin ring will be magnitude of field due to thin ring: k `dQ / (x^2 + r^2) * x / sqrt (x^2 + r^2) = 2 pi k r `dr * x / (x^2 + r^2)^(3/2). Summing over all such thin rings, which run from r = R1 to r = R2, we obtain the integral magnitude of field = integral ( 2 pi k r x /(x^2 + r^2)^(3/2) with respect to r, from R1 to R2). Evaluating the integral we find that magnitude of field = 2* pi k *x* | 1 /sqrt(x^2 + r1^2) - 1 / sqrt(x^2 + r2^2) | The direction of the field is along the x axis. If the point is close to the origin then x is close to 0 and x / sqrt(x^2 + r^2) is approximately equal to x / r, for any r much larger than x. This is because the derivative of x / sqrt(x^2 + r^2) with respect to x is r^2 / (x^2+r^2)^(3/2), which for x = 0 is just 1/r, having no x dependence. So at small displacement `dx from the origin the field strength will just be some constant multiple of `dx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Query 25 025. ********************************************* Question: `qQuery introductory set #1, 10-17 Explain how to find the potential difference in volts between two given points on the x axis, due to a given charge at the origin. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Work = integral k Q / x^2 between the two x values W = kQ*1Coulomb ( 1 / x1 - 1 / x2) potential difference = k * Q ( 1 / x1 - 1 / x2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Potential difference is the work per Coulomb of charge moved between the two points. To find this work you can multiply the average force on a Coulomb of charge by the displacement from the first point to the second. You can find an approximate average force by finding the force on a 1 Coulomb test charge at the two points and averaging the two forces. Multiplying this ave force by the displacement gives an approximate potential difference. Since the force is not a linear function of distance from the given charge, if the ratio of the two distances from the test charge is not small the approximation won't be particularly good. The approximation can be improved to any desired level of accuracy by partitioning the displacement between charges into smaller intervals of displacement and calculating the work done over each. The total work required is found by adding up the contributions from all the subintervals. University Physics students should understand how this process yields the exact work, which is the integral of the force function F(x) = k Q / x^2 between the two x values, yielding total work W = k * Q * 1 Coulomb ( 1 / x1 - 1 / x2) and potential difference V = k * Q ( 1 / x1 - 1 / x2). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qExplain how to find the potential difference between two points given the magnitude and direction of the uniform electric field between those points. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Work/Coulomb between the points = electric field * displacement V = E * `dr confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The work per Coulomb done between the two points is equal to the product of the electric field E and the displacement `dr. Thus for constant field E we have V = E * `dr. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qExplain how to find the average electric field between two points given a specific charge and the work done on the charge by the electric field as the charge moves between the points. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dV = dW / q E_ave = dV/ds confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You get ave force from work and distance: F_ave = `dW / `ds. You get ave electric field from work and charge: E_ave = F / q. An alternative: Find potential difference `dV = `dW / q. Ave electric field is Eave = `dV / `ds ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIn your own words explain the meaning of voltage. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It is distance between points * difference in electric field Confidence rating: 3
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Given Solution: `a** Voltage is the work done per unit of charge in moving charge from one point to another. ** STUDENT SOLUTION Voltage is the difference in electric field times the distance between two points: V = E*d The bigger the difference in voltage between two points, the greater potential to drive electrical current. INSTRUCTOR COMMENT Your answer was also correct. Note that the units for voltage are Joules / Coulomb, which is consistent with the given solution (work per unit of charge). However this unit can also be expressed as N * m / C , or (N / C) * m, consistent with your statement (N / C is the unit of electric field, so this would be the product of electric field and distance). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Query 26 026. Note that the solutions given here 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 switch between the two notations, since k = 1 / (4 pi epsilon0), or alternatively epsilon0 = 1 / (4 pi k). Introductory Problem Set 2 ********************************************* Question: `qBased on what you learned from Introductory Problem Set 2, how does the current in a wire of a given material, for a given voltage, depend on its length and cross-sectional area? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: E_wire = voltage/length_wire Increasing L decreases the field and inevitably the current. Increasing cross-sectional area increases the wire_vol and increases current confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The electric field in the wire is equal to the voltage divided by the length of the wire. So a longer wire has a lesser electric field, which results in less acceleration of the free charges (in this case the electrons in the conduction band), and therefore a lower average charge velocity and less current. The greater the cross-sectional area the greater the volume of wire in any given length, so the greater the number of charge carriers (in this case electrons), and the more charges to respond to the electric field. This results in a greater current, in proportion to the cross-sectional area. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow can the current in a wire be determined from the drift velocity of its charge carriers and the number of charge carriers per unit length? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: current = number of charges per unit length * drift velocity
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Given Solution: The charge carriers in a unit length will travel that length in a time determined by the average drift velocity. The higher the drift velocity the more quickly they will travel the unit length. This will result in a flow of current which is proportional to the drift velocity. Specifically if there are N charges in length interval `dL of the conductor and the drift velocity is v, all of the N charges will pass the end of the length interval in time interval `dt = `dL / v. The current can be defined as • current = # of charges passing a point / time required to pass the point Thus the current, in charges / unit of time passing the end of the length interval, is • current = N / `dt = N / (`dL / v) = (N / `dL) * v. N / `dL is the number of charges per unit length, and v is the drift velocity, so we can also say that • current = number of charges per unit length * drift velocity &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qWill a wire of given length and material have greater or lesser electrical resistance if its cross-sectional area is greater, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Greater CSA means greater current. If current is greater than it is implied the resistance will be lesser confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Greater cross-sectional area implies greater number of available charge carriers. For a given voltage and a given length of wire the electric field (equal to `dV / `dL) will be the same. Since it is the electric field that accelerates the charge carriers each charge will experience the same acceleration, independent of the cross-sectional area. The average drift velocity of the charge carriers will therefore be the same, regardless of the cross-sectional area. The result will be greater current for a given voltage. Greater current for a given voltage implies lesser electrical resistance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWill a wire of given material and cross-sectional area have greater or lesser electrical resistance if its length is greater, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: E = `dV / `dL greater length means lesser E which means less current flow which means greater electrical resistance confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The electric field is E = `dV / `dL, so greater length implies lesser electrical field for a given voltage, which implies less current flow. This implies greater electrical resistance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Principles and General Physics 16.24: Force on proton is 3.75 * 10^-14 N toward south. What is magnitude and direction of the field? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: direction is the same as the proton which indicates south based on its +charge E = F / q = 3.75 * 10^-14 N / (1.6 * 10^-19 C) = 2.36* 10^5 N / C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The direction of the electric field is the same as that as the force on a positive charge. The proton has a positive charge, so the direction of the field is to the south. The magnitude of the field is equal to the magnitude of the force divided by the charge. The charge on a proton is 1.6 * 10^-19 Coulombs. So the magnitude of the field is E = F / q = 3.75 * 10^-14 N / (1.6 * 10^-19 C) = 2.36* 10^5 N / C. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf the charges are represented by Q and -Q, what is the electric field at the midpoint? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: E_total = 2 k Q / r^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** This calls for a symbolic expression in terms of the symbol Q. The field from either charge is k Q / r^2, directed toward the negative charge. The field of both charges together is therefore E_total = 2 k Q / r^2, where r=.08 meters. ** STUDENT COMMENT: That is a tough one. I will have to read up on this one. I guess you just added the 2 because they are two charges? INSTRUCTOR RESPONSE: There are two charges and you are asked for the field at their midpoint. We find the field due to each of the two charges, then we add the two fields. Had the charges been of the same sign, rather than equal and opposite, the two fields would have been equal, but opposite, and would therefore have added up to zero. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: E = 9 * 10^9 N m^2 / C^2 * 33.0 * 10^-6 C / (.200 m)^2 = 7.43 * 10^6 N / C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A positive test charge Q at the given point will be repelled by the given positive charge, so will experience a force which is directly upward. The field has magnitude E = (k q Q / r^2) / Q, where q is the given charge and Q an arbitrary test charge introduced at the point in question. The field is the force per unit test charge, in this case (k q Q / r^2) / Q = k q / r^2. Substituting our given values we obtain E = 9 * 10^9 N m^2 / C^2 * 33.0 * 10^-6 C / (.200 m)^2 = 7.43 * 10^6 N / C. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery univ 22.34 / 22.32 11th edition 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Area =(.3 m)^2 = .09 m^2 S1 flux = (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-j) * .09 m^2 = 0. S2 flux = (-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. S3 flux = (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( j) * .09 m^2 = 0. S4 flux = (-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. S5 flux = (-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. S6 flux = (-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. S2 and S4, z = .3 m and z = 0 m so flux = .027 N m^2 / C and 0. S5 and S6, x = .3 m and x = 0 m so flux = -.045 N m^2 / C and 0. total flux = .027 N m^2 / C - .045 N m^2 / C = -.018 N m^2 / C. 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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: **** Advance correction of a common misconception: Flux is not a vector quantity so your flux values will not be multiples of the i, j and k vectors. The vectors normal 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. ** Self-critique (if necessary) ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery univ 22.47/22.45 11th edition 22.39 (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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For r that b < r < c the electric field is 4 pi k * 2q / r^2 = 8 pi k q / r^2 Field is zero for a < r < b and for c < r < d For any r that d < r the charge enclosed by the corresponding sphere = 6q of the two shells making the field: 4 pi k * 6q / r^2 = 24 pi k q / r^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ** STUDENT COMMENT with that said it now becomes a little wierd that the net charge would be 6q with a negative 2q on the outer surface of the inner shell and a -2q on the inner of the outer shell. i would think that these cancel each other, They really wouldnt cancel each other though because they dont have the same radius. I'll just trust that the field is 6q on the outside and that the charge is also k*6q/r^2 INSTRUCTOR RESPONSE The -2q of charge on the inside of the outer shell stays there, rather than migrating to the outside of the shell, because of the electric field created by the +2q charge on the inner shell. If the inner shell was removed, or the charge on it somehow neutralized, then the -2q on the inside of the outer shell would migrate to the outside of the shell, leaving 0 charge on the inside and charge +4q on the outside of that shell. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery univ 22.40 / 22.30 / 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: For r < a flux is: charge enclosed = 4pikL(alpha) electric field = flux / area = 4pikL (alpha) / (2pirL ) = 2k(alpha) / r. For a < r < b charge = 0 For b < r charge enclosed = 2L*alpha electric field = 4pik* 2L * alpha / (2pirL) = 4 k alpha / r. line charge + charge on inner cylinder + charge on outer cylinder = alpha * L alpha * L - alpha * L + charge on outer cylinder = alpha * L charge on outer cylinder = 2 alpha * L confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Self-critique Rating: ********************************************* Question: `qquery univ phy 13.64/23.62 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: Vab = E * r * ln(b/a) = 20,000 V/m * ln(1.8 cm / .0145 cm) * .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): ------------------------------------------------ Self-critique Rating: Query 27 027. ********************************************* Question: `qQuery Principles and General Physics 17.4: work by field on proton from potential +135 V to potential -55 V. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Delta potential = -55 V - (135 V) = -190 -190 V * 1.6 * 10^-19 C = -190 J / C * 1.6 * 10^-19 C = -3.0 * 10^-17 J. field does 3.0 * 10^-17 J of work on the charge. work = 180 volts * charge of 1 electron= 180 eV. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The change in potential is final potential - initial potential = -55 V - (135 V) = -190 V, so the change in the potential energy of the proton is -190 V * 1.6 * 10^-19 C = -190 J / C * 1.6 * 10^-19 C = -3.0 * 10^-17 J. In the absence of dissipative forces this is equal and opposite to the change in the KE of the proton; i.e., the proton would gain 3.09 * 10^-17 J of kinetic energy. Change in potential energy is equal and opposite to the work done by the field on the charge, so the field does 3.0 * 10^-17 J of work on the charge. Since the charge of the proton is equal in magnitude to that of an electron, he work in electron volts would be 180 volts * charge of 1 electron= 180 eV. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Principles and General Physics 17.8: Potential difference required to give He nucleus 65.0 keV of KE. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: He nuclues gains 2 eV KE To get 6.50 * 10^4 eV: voltage difference = .5( 6.50 * 10^4 volts) = 3.25 * 10^4 volts confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. 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: V = kq/r = 9.0*10^9N*m^2/C^2(1.60*10^-19C) / (2.5*10^-15m) = 5.8*10^5V 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 Work = PE=(1.60*10^-19C)(5.8*10^5V)= 9.2*10^-14 J. 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 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: Q = -1.2 * 10^-12 C 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 E from + infinity to .00065: (-1.08 * 10^-2 N m^2 / C) / (.00065 m) = -16.6 N m / C = -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): ------------------------------------------------ Self-critique Rating: 028. `Query 28 ********************************************* Question: `qQuery introductory problems set 54 #'s 1-7. Explain how to obtain the magnetic field due to a given current through a small current segment, and how the position of a point with respect to the segment affects the result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: B = k ' I L / r^2 * sin(theta) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** IL is the source. The law is basically an inverse square law and the angle theta between IL and the vector r from the source to the point also has an effect so that the field is B = k ' I L / r^2 * sin(theta). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery principles and general college physics problem 17.34: How much charge flows from each terminal of 7.00 microF capacitor when connected to 12.0 volt battery? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: C = Q / V Q = C * V= 7.00 microF * 12.0 volts = 7.00 microC / volt * 12.0 v = 84.0 microC. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Capacitance is stored charge per unit of voltage: C = Q / V. Thus the stored charge is Q = C * V, and the battery will have the effect of transferring charge of magnitude Q = C * V = 7.00 microF * 12.0 volts = 7.00 microC / volt * 12.0 volts = 84.0 microC of charge. This would be accomplished the the flow of 84.0 microC of positive charge from the positive terminal, or a flow of -84.0 microC of charge from the negative terminal. Conventional batteries in conventional circuits transfer negative charges. STUDENT COMMENT Ok. I didn’t really understand the +/- explanation though. INSTRUCTOR RESPONSE The positive terminal of a battery attracts negative charges, and/or repels positive charges. The negative terminal attracts positive charges, and/or repels negative charges. In a circuit where the available 'free charges' are negative, as in most circuits consisting of metal wires and various circuit elements. In such a circuit negative charges that reach the positive terminal are 'pumped' through the battery to the negative terminal (they wouldn't go there naturally; it takes energy to pull them away from the positive and get them to move to the negative terminal), where they are repelled. The result is a flow of negative charges toward the positive terminal, then away from the negative. This is completely equivalent to what would happen if the charge carriers were positive, moving in the opposite direction, away from the positive terminal and toward the negative. For a good time after circuits were put into use, nobody knew whether the charge carriers were positive or negative, or perhaps a mix of both. The convention prior to that time was that the direction of the current was the direction in which positive charge carriers would move (away from positive terminal, torward the negative). By the time the nature of the charges was discovered, the textbooks and engineering manuals had been around for awhile, and there was no way to change them. So the convention continues. STUDENT QUESTION I see that the unit is C not Farad? I understand the volt canceling out, but I thought capacity was measured 1 C/V? INSTRUCTOR RESPONSE Good question. The problem asked for the amount of charge. Charge is measured in Coulombs, abbreviated C. It's easy to confuse the C that stands for capacitance with the C that stands for Coulombs: • The unit C stands for Coulombs. • The variable C stands for capacitance. • To avoid confusion we have to be careful to keep the context straight. A Farad is a Coulomb per volt (C / V). So the unit of capacitance C is the Farad, or (C / V), where the C in the units stands for Coulombs. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: Explain how to obtain the magnetic field due to a circular loop at the center of the loop. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: B = sum(k ' I `dL / r^2 around entire loop Since I and r are constants: B = k ' I / r^2 sum(`dL). B = 2pirk ' I / r^2 = 2pik I / r. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For current running in a circular loop: Each small increment `dL of the loop is a source I `dL. The vector from `dL to the center of the loop has magnitude r, where r is the radius of the loop, and is perpendicular to the loop so the contribution of increment * `dL to the field is k ' I `dL / r^2 sin(90 deg) = I `dL / r^2, where r is the radius of the loop. The field is either upward or downward by the right-hand rule, depending on whether the current runs counterclockwise or clockwise, respectively. The field has this direction regardless of where the increment is located. The sum of the fields from all the increments therefore has magnitude B = sum(k ' I `dL / r^2), where the summation occurs around the entire loop. I and r are constants so the sum is B = k ' I / r^2 sum(`dL). The sum of all the length increments around the loop is the circumference 2 pi r of the loop so we have B = 2 pi r k ' I / r^2 = 2 pi k ' I / r. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery magnetic fields produced by electric currents. What evidence do we have that electric currents produce magnetic fields? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Ball being attracted towards the coil when current ran thru the wires confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: We do have evidence that electric currents produce magnetic fields. This is observed in engineering when laying current carrying wires next to each other. The current carrying wires produce magnetic fields that may affect other wires or possibly metal objects that are near them. This is evident in the video experiment. When Dave placed the metal ball near the coil of wires and turned the generator to produce current in the wires the ball moved toward the coil. This means that there was an attraction toward the coil which in this case was a magnetic field. INSTRUCTOR COMMENT: Good observations. A very specific observation that should be included is that a compass placed over a conducting strip or wire initially oriented in the North-South direction will be deflected toward the East-West direction. ** How is the direction of an electric current related to the direction of the magnetic field that results? ** GOOD STUDENT RESPONSE: The direction of the magnetic field relative to the direction of the electric current can be described using the right hand rule. This means simply using your right hand as a model you hold it so that your thumb is extended and your four fingers are flexed as if you were holding a cylinder. In this model, your thumb represents the direction of the electric current in the wire and the flexed fingers represent the direction of the magnetic field due to the current. In the case of the experiment the wire was in a coil so the magnetic field goes through the hole in the middle in one direction. ** Query problem 17.35 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat would be the area of a .20 F capacitor if plates are separated by 2.2 mm of air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: V = E * d C = Q / V = Q / (4 pi k Q / A * d) = A / (4 pi k d). A = 4 pi k d * C A = 4 pi * 9 * 10^9 N m^2 / C^2 * .20 C / V * .0022 m =5 * 10^7 N m^2 / C^2 * C / ( J / C) * m =5 * 10^7 m^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If a parallel-plate capacitor holds charge Q and the plates are separated by air, which has dielectric constant very close to 1, then the electric field between the plates is E = 4 pi k sigma = 4 pi k Q / A, obtained by applying Gauss' Law to each plate. The voltage between the plates is therefore V = E * d, where d is the separation. The capacitance is C = Q / V = Q / (4 pi k Q / A * d) = A / (4 pi k d). Solving this formula C = A / (4 pi k d) for A for the area we get A = 4 pi k d * C If capacitance is C = .20 F and the plates are separated by 2.2 mm = .0022 m we therefore have A = 4 pi k d * C = 4 pi * 9 * 10^9 N m^2 / C^2 * .20 C / V * .0022 m = 5 * 10^7 N m^2 / C^2 * C / ( J / C) * m = 5 * 10^7 N m^2 / (N m) * m = 5 * 10^7 m^2. ** STUDENT QUESTION I am not seeing where the 4pi k d came from... INSTRUCTOR RESPONSE 4 pi k Q / A * d is the same as 4 pi k d Q / A, by order of operations. So Q / (4 pi k Q / A * d) simplifies to A / (4 pi k d). The electric field near the surface of of a flat plate is 2 pi k * Q / A, as we find using the flux picture (total flux of charge Q is 4 pi k Q; a rectangular Gaussian surface and symmetry arguments are used to show that half the flux exits each end of the surface, resulting in field 2 pi k Q / A). The electric field between two oppositely charged plates is therefore 4 pi k * Q / A. Multiplying this field by the distance d between plates gives us the voltage. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 17.50 charge Q on capacitor; separation halved as dielectric with const K inserted; by what factor does the energy storage change? Compare the new electric field with the old. Note that the problem in the latest version of the text doubles rather than halves the separation. The solution for the halved separation, given here, should help you assess whether your solution was correct, and if not should help you construct a detailed self-critique. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: .5d would make V = E * d .5Voltage C C increases by 2 k stored energy .5 Q^2 / C decrease by 2k confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For a capacitor we know the following: • The electric field between the plates is 4 pi k Q / A (see solution to preceding problem), as long the separation d of the plates is small compared to the dimensions of the plates, and is independent of the separation. • Voltage is work / unit charge to move from one plate to the other. Since work = force * distance, work / unit charge is which is force / unit charge * distance between plates. Equivalently, since the electric field is the force per unit charge, work / unit charge is electric field * distance. That is, V = E * d. • Capacitance is Q / V, ratio of charge to voltage. • Energy stored is .5 Q^2 / C, which is just the work required to move charge Q across the plates with the 'average' voltage .5 Q / C (also obtained by integrating `dW = `dQ * V = `dq * q / C from q=0 to q = Q). • The dielectric increases capacitance by reducing the electric field, which thereby reduces the voltage between plates. The electric field will be 1 / k times as great, meaning 1/k times the voltage at any given separation. For the present situation we halve the separation of the plates and insert a dielectric with constant k. For a given Q, then, the electric field is fixed so that halving the separation d halves the voltage V = E * d. Halving the voltage V doubles the capacitance Q / V. Then inserting the dielectric reduces the field E, thereby reducing the voltage and increasing the capacitance by factor k. Thus the capacitance increases by factor 2 k. For given Q, this will decrease the stored energy .5 Q^2 / C by factor 2 k. ** STUDENT QUESTION: I am very confused on the correct answer. I assumed the voltage would stay constant. However, I think what the true answer is that voltage will increase by 1/k? My final answer is the same as yours, that the energy will increase by 2k. INSTRUCTOR RESPONSE: The capacitor is already charged, so Q remains constant. The effect of the dielectric is to decrease the electric field, which by itself would decrease the voltage to 1/k of its former value, which increases the capacitance by factor k. The effect of halving the distance is to decrease the voltage by another factor of 2, which increases the capacitance by factor 2. Since Q remains constant, the energy .5 Q^2 / C decreases by factor 2 k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery univ 24.48 / 24.50 (25.36 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery. What is the capacitance of this capacitor? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: E = 12 V / (4.7 mm) = 12 V / (.0047 m) = 2550 V / m 4 pi k sigma = 2250 V / m sigma = 2250 V / m / (4 pi k) = 2250 V / m / (4 pi * 9 * 10^9 N m^2 / C^2) =2.25*10^-8 C /m^2. Plate charge =.0256 m^2 * 2.25 * 10^-8 C / m^2 = 5.76 * 10^-10 C C = Q / V = 5.67 * 10^-10 C / (12 V) = 4.7 * 10^-11 C / V = 4.7 * 10^-11 Farads. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Fundamental principles include the fact that the electric field is very neary constant between parallel plates, the voltage is equal to field * separation, electric field from a single plate is 2 pi k sigma, the work required to displace a charge is equal to charge * ave voltage, and capacitance is charge / voltage. Using these principles we reason out the problem as follows: If the 4.7 mm separation experiences a 12 V potential difference then the electric field is E = 12 V / (4.7 mm) = 12 V / (.0047 m) = 2550 V / m, approx. Since the electric field of a plane charge distribution with density sigma is 2 pi k sigma, and since the electric field is created by two plates with equal opposite charge density, the field of the capacitor is 4 pi k sigma. So we have 4 pi k sigma = 2250 V / m and sigma = 2250 V / m / (4 pi k) = 2250 V / m / (4 pi * 9 * 10^9 N m^2 / C^2) = 2.25 * 10^-8 C / m^2. The area of the plate is .0256 m^2 so the charge on a plate is .0256 m^2 * 2.25 * 10^-8 C / m^2 = 5.76 * 10^-10 C. The capacitance is C = Q / V = 5.67 * 10^-10 C / (12 V) = 4.7 * 10^-11 C / V = 4.7 * 10^-11 Farads. The energy stored in the capacitor is equal to the work required to move this charge from one plate to another, starting with an initially uncharged capacitor. The work to move a charge Q across an average potential difference Vave is Vave * Q. Since the voltage across the capacitor increases linearly with charge the average voltage is half the final voltage, so we have vAve = V / 2, with V = 12 V. So the energy is energy = vAve * Q = 12 V / 2 * (5.76 * 10^-10 C) = 3.4 * 10^-9 V / m * C. Since the unit V / m * C is the same as J / C * C = J, we see that the energy is 3.4 * 10^-9 J. Pulling the plates twice as far apart while maintaining the same voltage would cut the electric field in half (the voltage is the same but charge has to move twice as far). This would imply half the charge density, half the charge and therefore half the capacitance. Since we are moving only half the charge through the same average potential difference we use only 1/2 the energy. Note that the work to move charge `dq across the capacitor when the charge on the capacitor is `dq * V = `dq * (q / C), so to obtain the work required to charge the capacitor we integrate q / C with respect to q from q = 0 to q = Q, where Q is the final charge. The antiderivative is q^2 / ( 2 C ) so the definite integral is Q^2 / ( 2 C). This is the same result obtained using average voltage and charge, which yields V / 2 * Q = (Q / C) / 2 * Q = Q^2 / (2 C) Integration is necessary for cylindrical and spherical capacitors and other capacitors which are not in a parallel-plate configuration. ** query univ 24.49/24.51 (25.37 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery. If the battery remains connected and plates are pulled to separation 9.4 mm then what are the capacitance, charge of each plate, electric field, energy stored? The potential difference between the plates is originally 12 volts. 12 volts over a 4.7 mm separation implies electric field = potential gradient = 12 V / (.0047 m) = 2500 J / m = 2500 N / C, approx.. The electric field is E = 4 pi k * sigma = 4 pi k * Q / A so we have Q = E * A / ( 4 pi k) = 2500 N / C * (.16 m)^2 / (4 * pi * 9 * 10^9 N m^2 / C^2) = 5.7 * 10^-7 N / C * m^2 / (N m^2 / C^2) = 5.7 * 10^-10 C, approx.. The energy stored is E = 1/2 Q V = 1/2 * 5.6 * 10^-10 C * 12 J / C = 3.36 * 10^-9 J. If the battery remains connected then if the plate separation is doubled the voltage will remain the same, while the potential gradient and hence the field will be halved. This will halve the charge on the plates, giving us half the capacitance. So we end up with a charge of about 2.8 * 10^-10 C, and a field of about 1250 N / C. The energy stored will also be halved, since V remains the same but Q is halved. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery univ (not present in 13th edition) 24.68 (25.52 10th edition). A solid conducting sphere radius of radius R carries charge Q. What is the electric-field energy density at distance r < R from the center of the sphere? What is the electric-field energy density at distance r > R from the center of the sphere? What is the total energy in the field of this sphere? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Energy density = .5 C V^2 / (vol) = .5 C V^2 / (d * A) Energy density = k = 4 pi / epsilon0 Energy density = 1 / (8 pi k) E^2. r > R E = Q / (4 pi epsilon0 r^2) Energy density = .5 epsilon0 E^2 = .5 epsilon0 Q^2 / (16 pi^2 epsilon0^2 r^4) = Q^2 / (32 pi^2 epsilon0 r^4) Energy in shell = energy density * volume = Q^2 / (32 pi^2 epsilon0 r^4) * 4 pi r^2 `dr = Q^2 / (8 pi epsilon0 r^2) `dr. Integrating get: Total energy = Q^2 / (8 pi epsilon0 R) confidence rating #$&*:ad to view solution for greater understanding ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We will find the energy density function then integrate that density function over all of space to find the total energy of the distribution. We will compare this with the energy required to assemble the distribution, and will find that the two are equal. • To integrate the energy density over all space we will find the total energy in a thin spherical shell of radius r and thickness `dr, then use this result to obtain our integral. • Then we will integrate to find the work required to assemble the charge on the surface of the sphere. Energy density, defined by dividing the energy .5 C V^2 required to charge a parallel-plate capacitor by the volume occupied by its electric field, is • Energy density = .5 C V^2 / (volume) = .5 C V^2 / (d * A), where d is the separation of the plates and A the area of the plates. Since C = epsilon0 A / d and V = E * d this gives us .5 epsilon0 A / d * (E * d)^2 / (d * A) = .5 epsilon0 E^2 so that • Energy density = .5 epsilon0 E^2, or in terms of k = 4 pi / epsilon0 • Energy density = 1 / (8 pi k) E^2. Since your text uses epsilon0 I'll do the same on this problem. In this problem the epsilon0 notation makes a good deal of sense: For the charged sphere the electric field for r < R is zero, since there can be no electric field inside a conductor. For r > R the electric field is • E = Q / (4 pi epsilon0 r^2), and therefore • energy density = .5 epsilon0 E^2 = .5 epsilon0 Q^2 / (16 pi^2 epsilon0^2 r^4) = Q^2 / (32 pi^2 epsilon0 r^4). The energy density (i.e., the energy per unit of volume) between r and r + `dr is nearly constant if `dr is small. As we saw above the energy density will be approximately Q^2 / (32 pi^2 epsilon0 r^4). The volume of space between r and r + `dr is approximately A * `dr = 4 pi r^2 `dr. The expression for the energy lying in the shell between distance r and r + `dr is therefore approximately • energy in shell = energy density * volume = Q^2 / (32 pi^2 epsilon0 r^4) * 4 pi r^2 `dr = Q^2 / (8 pi epsilon0 r^2) `dr. This leads to a Riemann sum over radius r. As we let `dr approach zero the Riemann sum approaches an integral with integrand Q^2 / (8 pi epsilon0 r^2), integrated with respect to r. To get the energy between two radii we therefore integrate this expression between those two radii. If we integrate this expression from r = R to infinity we get the total energy of the field of the charged sphere. This integral gives us • total energy = Q^2 / (8 pi epsilon0 R) (alternatively using k = 4 pi k / epsilon0, this result would be k Q^2 / (2 R). This form is less complicated to the eye and will be used in the comparison below.) We compare this with the work required to charge the sphere: • To bring a charge `dq from infinity to a sphere containing charge q requires work k q / R `dq • To charge the entire sphere the work would therefore be the integral of k q / R with respect to q. • We integrate from q = 0 to q = Q, obtaining the total work required to charge the sphere. Our antiderivative is k (q^2 / 2) / r. If we evaluate this antiderivative at lower limit 0 and upper limit Q we get the total work, which is k Q^2 / (2 R). This agrees with our previous result, obtained by integrating the energy density of the field. Since k = 1 / (4 pi epsilon0) the work is k Q^2 / (2 R) = Q^2 / (8 pi epsilon0 R). So the energy in the field is equal to the work required to assemble the charge distribution. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 029. `Query 29 ********************************************* Question: `qQuery introductory problem set 54 #'s 8-13 Explain how to determine the magnetic flux of a uniform magnetic field through a plane loop of wire, and explain how the direction of the field and the direction of a line perpendicular to the plane of the region affect the result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: multiply the area of the loop by the strength of the electric field confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To do this we need to simply find the area of the plane loop of wire. If we are given the radius we can find the area using Pi * r ^2 Then we multiply the area of the loop (In square meters ) by the strength of the field (in tesla). This will give us the strength of the flux if the plane of the loop is perpendicular to the field. If the perpendicular to the loop is at some nonzero angle with the field, then we multiply the previous result by the cosine of the angle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qExplain how to determine the average rate of change of magnetic flux due to a uniform magnetic field through a plane loop of wire, as the loop is rotated in a given time interval from an orientation perpendicular to the magnetic field to an orientation parallel to the magnetic field. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: multiply the area by testla to find the flux when the loop is perpendicular to the field. When parallel its zero confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** EXPLANATION BY STUDENT: The first thing that we need to do is again use Pi * r ^ 2 to find the area of the loop. Then we multiply the area of the loop (m^2) by the strength of the field (testla) to find the flux when the loop is perpendicular to the field. Then we do the same thing for when the loop is parallel to the field, and since the cos of zero degrees is zero, the flux when the loop is parallel to the field is zero. This makes sense because at this orientation the loop will pick up none of the magnetic field. So now we have Flux 1 and Flux 2 being when the loop is perpendicular and parallel, respectively. So if we subtract Flux 2 from flux 1 and divide this value by the given time in seconds, we will have the average rate of change of magnetic flux. If we use MKS units this value will be in Tesla m^2 / sec = volts. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qExplain how alternating current is produced by rotating a coil of wire with respect to a uniform magnetic field. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: A changing magnetic flux is generated which produces voltage which produces current confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: Y ou rotate a coil of wire end over end inside a uniform magnetic field. When the coil is parallel to the magnetic field, then there is no magnetic flux, and the current will be zero. But then when the coil is perpendicular to the field or at 90 degrees to the field then the flux will be strongest and the current will be moving in one direction. Then when the coil is parallel again at 180 degrees then the flux and the current will be zero. Then when the coil is perpendicular again at 270 degrees, then the flux will be at its strongest again but it will be in the opposite direction as when the coil was at 90 degrees. So therefore at 90 degrees the current will be moving in one direction and at 270 degrees the current will be moving with the same magnitude but in the opposite direction. COMMENT: Good. The changing magnetic flux produces voltage, which in turn produces current. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Principles and General College Physics 18.04. 120V toaster with 4.2 amp current. What is the resistance? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: R = V / I = 120V / 4.2A = 28.57 ohms confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: current = voltage / resistance (Ohm's Law). The common sense of this is that for a given voltage, less resistance implies greater current while for given resistance, greater voltage implies greater current. More specifically, current is directly proportional to voltage and inversely proportional to resistance. In symbols this relationship is expressed as I = V / R. In this case we know the current and the voltage and wish to find the resistance. Simple algebra gives us R = V / I. Substituting our known current and voltage we obtain R = 120 volts / 4.2 amps = 29 ohms, approximately. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Principles and General College Physics 18.28. Max instantaneous voltage to a 2.7 kOhm resistor rated at 1/4 watt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: P = V^2 / R 0.25 = V^2 / 2700 V = 26v confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Voltage is energy per unit of charge, measured in Joules / Coulomb. Current is charge / unit of time, measured in amps or Coulombs / second. Power is energy / unit of time measured in Joules / second. The three are related in a way that is obvious from the meanings of the terms. If we multiply Joules / Coulomb by Coulombs / second we get Joules / second, so voltage * current = power. In symbols this is power = V * I. Ohm's Law tells us that current = voltage / resistance.In symbols this is I = V / R. So our power relationship power = V * I can be written power = V * V / R = V^2 / R. Using this relationship we find that V = sqrt(power * R), so in this case the maximum voltage (which will produce the 1/4 watt maximum power) will be V = sqrt(1/4 watt * 2.7 * 10^3 ohms) = 26 volts. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qQuery general college physics problem 18.39; compare power loss if 520 kW delivered at 50kV as opposed to 12 kV thru 3 ohm resistance.** The current will not be the same at both voltages. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: It is important to understand that power (J / s) is the product of current (C / s) and voltage (J / C). So the current at 50 kV kW will be less than 1/4 the current at 12 kV. To deliver 520 kW = 520,000 J / s at 50 kV = 50,000 J / C requires current I = 520,000 J/s / (50,000 J/C) = 10.4 amps. This demonstrates the meaning of the formula P = I V. To deliver 520 kW = 520,000 J / s at 12 kV = 12,000 J / C requires current I = 520,000 J/s / (12,000 J/C) = 43.3 amps. The voltage drops through the 3 Ohm resistance will be calculated as the product of the current and the resistance, V = I * R: The 10.4 amp current will result in a voltage drop of 10.4 amp * 3 ohms = 31.2 volts. The 43.3 amp current will result in a voltage drop of 40.3 amp * 3 ohms = 130 volts. The power loss through the transmission wire is the product of the voltage ( J / C ) and the current (J / S) so we obtain power losses as follows: At 520 kV the power loss is 31.2 J / C * 10.4 C / s = 325 watts, approx. At 12 kV the power loss is 130 J / C * 43.3 C / s = 6500 watts, approx. Note that the power loss in the transmission wire is not equal to the power delivered by the circuit, which is lost through a number of parallel connections to individual homes, businesses, etc.. The entire analysis can be done by simple formulas but without completely understanding the meaning of voltage, current, resistance, power and their relationships it is very easy to get the wrong quantities in the wrong places, and especially to confuse the power delivered with the power loss. The analysis boils down to this: I = P / V, where P is the power delivered. Ploss = I^2 R, where R is the resistance of the circuit and Ploss is the power loss of the circuit. So Ploss = I^2 * R = (P/V)^2 * R = P^2 * R / V^2. This shows that power loss across a fixed resistance is inversely proportional to square of the voltage. So that the final voltage, which is less than 1/4 the original voltage, implies more than 16 times the power loss. A quicker solution through proportionalities: For any given resistance power loss is proportional to the square of the current. For given power delivery current is inversely proportional to voltage. So power loss is proportional to the inverse square of the voltage. In this case the voltage ratio is 50 kV / (12 kV) = 4.17 approx., so the ratio of power losses is about 1 / 4.17^2 = 1 / 16.5 = .06. Note that this is the same approximate ratio you would get if you divided your 324.5 watts by 5624.7 watts. ** STUDENT COMMENT This is confusing. I will have to keep reading over the problem until I understand the solution….I’m afraid that I will have problems with this test…we will see……. INSTRUCTOR RESPONSE The concept of thinking these ideas through in terms of the units is developed in the introductory problem sets, and may be helpful. A key idea here is that power (which you should understand is energy / time interval, measured in Joules/second or watts) is the product of current (in C / sec or amps) and voltage (in J / C or volts). If you multiply the number of Coulombs per second, by the number of Joules per Coulomb, you get the number of Joules per second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery univ students with 12th and later editions should attempt this problem before reading the solution; 11th edition 25.62 (26.50 10th edition) A rectangular block of a homogeneous material (i.e., with constant resistivity throughout) has dimensions d x 2d x 3d. We have a source of potential difference V. To which faces of the solid should the voltage be applied to attain maximum current density and what is the density? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: R = rho * L / A = rho * (3d) / (2 d^2) = 3 / 2 rho / d I = V / R = V / (3/2 rho / d) = 2d V / (3 rho) Current density is I / A = (2 d V / (3 rho) ) / (2 d^2) = V / (3 rho d) = 1/3 V / (rho d) For d x 3d: R = rho * L / A = rho * (2d) / (3 d^2) = 2 / 3 rho / d I = V / R = V / (2/3 rho / d) = 3 d V / (2 rho) Current density is I / A = (3 d V / (2 rho) ) / (3 d^2) = V / (2 rho d) = 1/2 V / (rho d). For 3d x 2d: R = rho * L / A = rho * (d) / (6 d^2) = 1/6 rho / d I = V / R = V / (1/6 rho / d) = 6 d V / (rho) Current density is I / A = (6 d V / (rho) ) / (6 d^2) = V / (rho d) = V / (rho d) Max current density happens on the largest face confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** First note that the current I is different for diferent faces. The resistance of the block is proportional to the distance between faces and inversely proportional to the area, so current is proportional to the area and inversely proportional to the distance between faces. Current density is proportional to current and inversely proportional to the area of the face, so current density is proportional to area and inversely proportional to the distance between faces and to area, leaving current inversely proportional to distance between faces. For the faces measuring d x 2d we have resistance R = rho * L / A = rho * (3d) / (2 d^2) = 3 / 2 rho / d so current is I = V / R = V / (3/2 rho / d) = 2d V / (3 rho). Current density is I / A = (2 d V / (3 rho) ) / (2 d^2) = V / (3 rho d) = 1/3 V / (rho d). For the faces measuring d x 3d we have resistance R = rho * L / A = rho * (2d) / (3 d^2) = 2 / 3 rho / d so current is I = V / R = V / (2/3 rho / d) = 3 d V / (2 rho). Current density is I / A = (3 d V / (2 rho) ) / (3 d^2) = V / (2 rho d) = 1/2 V / (rho d). For the faces measuring 3d x 2d we have resistance R = rho * L / A = rho * (d) / (6 d^2) = 1 / 6 rho / d so current is I = V / R = V / (1/6 rho / d) = 6 d V / (rho). Current density is I / A = (6 d V / (rho) ) / (6 d^2) = V / (rho d) = V / (rho d). Max current density therefore occurs when the voltage is applied to the largest face. ** STUDENT QUESTION I don’t really understand why current isn’t constant throughout the object. I can follow the solution, but I don’t understand that one point. My solution was based on the constant current which I see is not correct. INSTRUCTOR RESPONSE If the voltage is applied to different faces then you will get different currents. The potential gradient is greater if the faces are closer together. This effect would tend to result in greater current for faces that are closer together. If the faces are closer together the areas of the faces will be greater, so the cross-sectional area will be greater. This effect would result in greater current. So the faces closest to together will experience the greatest potential gradient, and result in the greatest cross-sectional area, resulting in the greatest current. However the question asks about current density, not current. Intuitively, we know that the potential gradient is the electric field responsible for accelerating the charges, so that the charge density will just be proportional to the potential gradient, so that if the voltage is applied to the faces with the least separation the current density will be the greatest. In fact, since the separations are d, 2d and 3d we can see that the potential gradients will be V / d, 1/2 V / d and 1/3 V / d so the other two current densities will be 1/2 and 1/3 as great as the maximum. This can be worked out symbolically for the general case: R = rho * L / A, and current is I = V / R = V * A / (rho L) so current density is I / A = V / (rho L) The greatest current density occurs for the least value of L, which for the given situation is d, the separation between the largest faces. Working out the details more fully is be unnecessary but might be instructive: The faces closest together are separated by distance d, and have cross-sectional area 2 d * 3 d = 6 d^2. So the resistance is R = rho * L / A = rho * d / (6 d^2) = rho / (6 d) = 1/6 * rho / d The faces furthest apart are separated by distance 3d, and have cross-sectional area d * 2 d = 2 d^2. So the resistance is R = rho * L / A = rho * d / (2 d^2) = rho / (2 d) = 1/2 * rho / d Similar analysis shows that the resistance between faces separated by 2 d is rho / (3/2 d) = 2/3 * rho / d. Thus the currents would be I = V / R = 6 V * d / rho for the closest faces I = V / R = 2 V d / rho for the most widely separate faces and I = V / R = 2 /3 V d / rho for the in-between separation. The respective current densities, current / area, would be (6 V d / rho) / (6 d^2) = V / (rho d) (2 V d / rho) / (3 d^2) = 2/3 V / (rho d) (2/3 V d / rho) / (2d^2) = 1/3 V / (rho d). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 030. `Query 30 ********************************************* Question: `qQuery introductory problem set 54 #'s 14-18. Explain whether, and if so how, the force on a charged particle due to the field between two capacitor plates is affected by its velocity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Force exerted on the charged particle by an electric field does not depend on its velocity confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** There is a force due to the electric field between the plates, but the effect of an electric field does not depend on velocity. The plates of a capacitor do not create a magnetic field. ** Explain whether, and if so how, the force on a charged particle due to the magnetic field created by a wire coil is affected by its velocity. ** A wire coil does create a magnetic field perpendicular to the plane of the coil. If the charged particle moves in a direction perpendicular to the coil then a force F = q v B is exerted by the field perpendicular to both the motion of the particle and the direction of the field. The precise direction is determined by the right-hand rule. ** Explain how the net force changes with velocity as a charged particle passes through the field between two capacitor plates, moving perpendicular to the constant electric field, in the presence of a constant magnetic field oriented perpendicular to both the velocity of the particle and the field of the capacitor. ** At low enough velocities the magnetic force F = q v B is smaller in magnitude than the electrostatic force F = q E. At high enough velocities the magnetic force is greater in magnitude than the electrostatic force. At a certain specific velocity, which turns out to be v = E / B, the magnitudes of the two forces are equal. If the perpendicular magnetic and electric fields exert forces in opposite directions on the charged particle then when the magnitudes of the forces are equal the net force on the particle is zero and it passes through the region undeflected. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: `qquery univ 27.60 / 27.56 11th edition 27.60 (28.46 10th edition). cyclotron 3.5 T field. What is the radius of orbit for a proton with kinetic energy 2.7 MeV? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: K = 2.7 MeV = 2.7x10^6 eV (2.7x10^6 eV)(1.6x10^-19 J / eV) = 4.32x10^-13 J v = sqrt( 2 * K / m) v = sqrt(2 (4.32x10^-13 J) / (1.67x10^-27 kg)) = 2.27x10^7 m/s R = (mv)/(qB) R = [(1.67x10^-27 kg)(2.27x10^7 m/s)] / [(1.6x10^-19 C)(3.5 T)] = 0.068 m omega = v/R = (2.27x10^7 m/s) / (0.068 m) = 3.34x10^8 rad/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We know that the centripetal force for an object moving in a circle is F = m v^2 / r. In a magnetic field perpendicular to the velocity this force is equal to the magnetic force F = q v B. So we have m v^2 / r = q v B so that r = m v / (q B). A proton with ke 2.7 MeV = 2.7 * 10^6 * (1.6 * 10^-19 J) = 4.32 * 10^-13 J has velocity such that v = sqrt(2 KE / m) = sqrt(2 * 3.2 * 10^-13 J / (1.67 * 10^-27 kg) ) = 2.3 * 10^7 m/s approx.. So we have r = m v / (q B) = 1.67 * 10^-27 kg * 2.3 * 10^7 m/s / (1.6 * 10^-19 C * 3.5 T) = .067 m approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the radius of orbit for a proton with kinetic energy 5.4 MeV? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: K = 5.4 MeV = 5.4x10^6 eV (5.4x10^6 eV)(1.6x10^-19 J / eV) = 8.64x10^-13 J v = sqrt( 2 * K / m) v = sqrt(2 *(8.64x10^-13 J) / (1.67x10^-27 kg)) = 3.22x10^7 m/s R = (mv)/(qB) R = [(1.67x10^-27 kg)(3.22x10^7 m/s)] / [(1.6x10^-19 C)(3.5 T)] = 0.096 m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Doubling the KE of the proton increases velocity by factor sqrt(2) and therefore increases the radius of the orbit by the same factor. We end up with a radius of about .096 m. ** STUDENT QUESTION what numbers were used to find this? INSTRUCTOR RESPONSE In Problem 27.60, above, we found the radius of orbit for a proton with kinetic energy 2.7 MeV. Here we are finding the radius for a proton with twice the KE. We could do this in the same manner as before, and we would get the same result. However thinking in terms of the proportionality, as is done here, is both more efficient and more instructive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery univ 27.74 / 27.72 11th edition 28.66 (was 28.52) rail gun bar mass m with current I across rails, magnetic field B perpendicular to loop formed by bars and rails What is the expression for the magnitude of the force on the bar, and what is the direction of the force? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Part I: Force = ILB Part II: v_x^2 = v0^2 + 2a_x(x - x0) v^2 = 2ad d = v^2 / (2a) d = (v^2 * m) / (2ILB) Part III: d = [(1.12x10^4 m/s)^2 * (25 kg)] / [(2 *(2000 A)(0.50 m)(0.50 T)] d = 3.14x10^6 m = 3140 km confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The length of the bar is given as L. So the force is I L B, since the current and field are perpendicular. The acceleration of the bar is therefore a = I L B / m. If the distance required to achieve a given velocity is `ds and initial velocity is 0 then vf^2 = v0^2 + 2 a `ds gives us ds = (vf^2 - v0^2) / (2 a) = vf^2 / (2 a). If v stands for the desired final velocity this is written `ds = v^2 / (2 a). In terms of I, L, B and m we have `ds = v^2 / (2 I L B / m) = m v^2 / (2 I L B). Note that we would get the same expression using KE: since (neglecting dissipative losses) we have `dKE = `dW = F `ds we have `ds = `dKE / F = 1/2 m v^2 / (I L B). For the given quantities we get `ds = 25 kg * (1.12 * 10^4 m/s)^2 / (2 * 2000 amps * .50 Tesla * .5 meters) = 3.2 * 10^6 meters, or about 3200 km. ** STUDENT QUESTION I could find the magnitude of the force, but I don’t really understand the velocity equation laid out above. INSTRUCTOR RESPONSE vf^2 = v0^2 + 2 a `ds is one of the basic equations of uniformly accelerated motion, easily derived from the definitions of velocity and acceleration: vAve = (v0 + vf) / 2 = `ds / `dt aAve = `dv / `dt = (vf - v0) / `dt. We get the equations `ds / `dt = (v0 + vf) / 2a = (vf - v0) / `dt from which we can eliminate `dt, obtaining vf^2 = v0^2 + 2 a `ds. Here, `ds stands for the displacement, v0 and vf for initial and final velocities, and a for acceleration on the interval in question. Armed with this equation, then, and having found the expression for a, knowing that the initial velocity is zero we easily solve for the `ds necessary to achive the desired vf. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery 28.66 u quark + 2/3 e and d quark -1/3 e counterclockwise, clockwise in neutron (r = 1.20 * 10^-15 m) What are the current and the magnetic moment produced by the u quark? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I A = q v / (2 pi r) * pi r^2 = q v r / 2. total magnetic moment= 2/3 e * v r / 2 + 2 ( 1/3 e * v r / 2) = 4/3 e v r / 2 = 2/3 e v r 2/3 e v r = mu v = 3/2 mu / (e r) = 3/2 * 9.66 * 10^-27 A m^2 / (1.6 * 10^-19 C * 1.20 * 10^-15 m) = 7.5 * 10^7 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 Clearer after viewing solution
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Given Solution: ** If r is the radius of the orbit and v the velocity then the frequency of an orbit is f = v / (2 pi r). The frequency tells you how many times the charge passes a given point per unit of time. If the charge is q then the current must therefore be I = q f = q v / (2 pi r). Half the magnetic moment is due to the u quark, which carries charge equal and opposite to the combined charge of both d quarks, the other half to the d quarks (which circulate, according to this model, in the opposite direction with the same radius so that the two d quarks contribute current equal to, and of the same sign, as the u quark). The area enclosed by the path is pi r^2, so that the magnetic moment of a quark is I A = q v / (2 pi r) * pi r^2 = q v r / 2. The total magnetic moment is therefore 2/3 e * v r / 2 + 2 ( 1/3 e * v r / 2) = 4/3 e v r / 2 = 2/3 e v r.. Setting this equal to the observed magnetic moment mu we have 2/3 e v r = mu so that v = 3/2 mu / (e r) = 3/2 * 9.66 * 10^-27 A m^2 / (1.6 * 10^-19 C * 1.20 * 10^-15 m) = 7.5 * 10^7 m/s, approx.. Note that units are A m^2 / (C m) = C / s * m^2 / (C m) = m / s. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery univ 28.78 / 28.70 11th edition 28.68 (29.56 10th edition) infinite L-shaped conductor toward left and downward. Point a units to right of L along line of current from left. Current I. What is the magnetic field at the specified point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: B = (‘mu0 * I) / (2*pi*R)***from book B = (1/2) * [(‘mu0 * I) / (2*pi*R)] B = (mu0 * I) / (4*pi*R) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE FOLLOWED BY SOLUTION: I could not figure out the magnetic field affecting point P. the current is cursing ** I assume you mean 'coursing', though the slip is understandable ** toward P then suddnely turns down at a right angle. If I assume that the magnetic field of a thin wire is radial in all directions perpendicular to the wire, then it is possible that at least one field line would be a straight line from the wire to point P. It seems to me that from that field line,down the to the lower length of the wire, would affect at P. SOLUTION: The r vector from any segment along the horizontal section of the wire would be parallel to the current segment, so sin(theta) would be 0 and the contribution `dB = k ' I `dL / r^2 sin(theta) would be zero. So the horizontal section contributes no current at the point. Let the y axis be directed upward with its origin at the 'bend'. Then a segment of length `dy at position y will lie at distance r = sqrt(y^2 + a^2) from the point and the sine of the angle from the r vector to the point is a / sqrt(y^2 + a^2). The field resulting from this segment is therefore `dB = k ' I `dy / (a^2 + y^2) * a / sqrt(a^2 + y^2). Crossing the I `dy vector with the r vector tells us the `dB is coming at us out of the paper (fingers extended along neg y axis, ready to 'turn' toward r results in thumb pointing up toward us away from the paper). This is the direction for all `dB contributions so B will have the same direction. Summing all contributions we have sum(k ' I `dy / (a^2 + y^2) * a / sqrt(a^2 + y^2), y from 0 to -infinity). Taking the limit as `dy -> 0 we get the integral of k ' I a / (a^2 + y^2)^(3/2) with respect to y, with y from 0 to -infinity. This integral is -k ' I / a. So the field is B = - k ' I / a, directed upward out of the page. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 031. `Query 31 ********************************************* Question: `quniv query 29.54 (30.36 10th edition) univ upward current I in wire, increasing at rate di/dt. Loop of height L, vert sides at dist a and b from wire. When the current is I what is the magnitude of B at distance r from the wire and what is the magnetic flux through a strip at this position having width `dr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Went through the given solution A = L * `dr magnetic flux = 2 k ' I / r * (L `dr). total magnetic field : sum(2 k ' I / r * L `dr, r from a to b). flux = 2 k ' L ln | b / a | * I rate of change of flux = 2 k ' L ln | b / a | * dI/dt. confidence rating #$&*:t through the given solution ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The magnetic field due to the wire at distance r is 2 k ' I / r. The field is radial around the wire and so by the right-hand rule (thumb in direction of current, fingers point in direction of field) is downward into the page. The area of the strip is L * `dr. The magnetic flux thru the strip is therefore 2 k ' I / r * (L `dr). The total magnetic field over a series of such strips partitioning the area is thus sum(2 k ' I / r * L `dr, r from a to b). Taking the limit as `dr -> 0 we get integral (2 k ' I / r * L with respect to r, r from a to b). Our antiderivative is 2 k ' I ln | r | * L; the definite integral therefore comes out to flux = 2 k ' L ln | b / a | * I. If I is changing then we have rate of change of flux = 2 k ' L ln | b / a | * dI/dt. This is the induced emf through a single turn. You can easily substitute a = 12.0 cm = .12 m, b = 36.0 cm = .36 m, L = 24.0 cm = .24 m and di/dt = 9.60 A / s, and multiply by the number of turns. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: A 320-loop square coil 21 cm on a side rotates about an axis perpendicular to a .65 T mag field. What frequency of oscillation will produce a peak 120-v output? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Flux = B*A*n, where n is the number of loops: Flux = (0.65 T)*(.21 m)^2*(320) = 9.17 T*m^2 phi(t) = 9.173 T m^2 * sin(2 pi f t). V(t) = phi ' (t) = 9.173 T m^2 * 2 pi f cos(2 pi f t). Maximum voltage occurs when cos(2 pi f t) = 1. Max voltage = 9.173 T m^2 * 2 pi f. 9.173 T m^2 * 2 pi f = 120 V f = 120 V / (9.173 T m^2 * 2 pi) = 2.08 V / / (T m^2) = 2.08 T m^2 / s / (T m^2) = 2.08 s^-1. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: I wouldn't advocate using a formula from the book to solve this problem. Common sense, starting from the premise that the voltage function is the derivative of the flux function, is much more efficient (way fewer formulas to remember and less chance of using the wrong one). The maximum flux is .65 T * (.21 m)^2/ loop * 320 loops = 9.173 T m^2. So the flux as a function of clock time could be modeled by • phi(t) = 9.173 T m^2 * sin(2 pi f t). The voltage induced by changing flux is the rate of change of flux with respect to clock time. So the voltage function is the t derivative of the flux: • V(t) = phi ' (t) = 9.173 T m^2 * 2 pi f cos(2 pi f t). Maximum voltage occurs when cos(2 pi f t) = 1. At this instant the voltage is • max voltage = 9.173 T m^2 * 2 pi f. Setting this equal to the peak voltage we get • 9.173 T m^2 * 2 pi f = 120 V so that • f = 120 V / (9.173 T m^2 * 2 pi) = 2.08 V / / (T m^2) = 2.08 T m^2 / s / (T m^2) = 2.08 s^-1. We can generalize this symbolically by replacing 9.173 T m^2 by phi_max, which represents the maximum flux. So a generator with maximum flux phi_max, rotating a frequency f has flux function • phi(t) = phi_max cos(2 pi f t) with t derivative • V(t) = phi ' (t) = phi_max cos(2 pi f t). Everything follows easily from this formulation, with no need to memorize the formulas that result. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&! " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!#*&!