#$&* course Phy 242 7/19 5pm 026. Query 27
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Given Solution: The electric field in the wire is equal to the voltage divided by the length of the wire. So a longer wire has a lesser electric field, which results in less acceleration of the free charges (in this case the electrons in the conduction band), and therefore a lower average charge velocity and less current. The greater the cross-sectional area the greater the volume of wire in any given length, so the greater the number of charge carriers (in this case electrons), and the more charges to respond to the electric field. This results in a greater current, in proportion to the cross-sectional area. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qHow can the current in a wire be determined from the drift velocity of its charge carriers and the number of charge carriers per unit length? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I = Q / t I = N / 'dt I = N / ('dL/v) I = (N / 'dL) * v v is drift velocity N/ 'dL is # charges per unit length Therefore: Current = Number of charge carriers per unit length * drift velocity confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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 area = greater # charge carriers = greater current = less resistance confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Greater cross-sectional area implies greater number of available charge carriers. For a given voltage and a given length of wire the electric field (equal to `dV / `dL) will be the same. Since it is the electric field that accelerates the charge carriers each charge will experience the same acceleration, independent of the cross-sectional area. The average drift velocity of the charge carriers will therefore be the same, regardless of the cross-sectional area. The result will be greater current for a given voltage. Greater current for a given voltage implies lesser electrical resistance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qWill a wire of given material and cross-sectional area have greater or lesser electrical resistance if its length is greater, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Greater length = lesser electrical field = less current flow = greater resistance confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: E = F / q E = 3.75 x 10^-14 / 1.6 x 10^-19 E = 2.3 x 10^5 N/C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The direction of the electric field is the same as that as the force on a positive charge. The proton has a positive charge, so the direction of the field is to the south. The magnitude of the field is equal to the magnitude of the force divided by the charge. The charge on a proton is 1.6 * 10^-19 Coulombs. So the magnitude of the field is E = F / q = 3.75 * 10^-14 N / (1.6 * 10^-19 C) = 2.36* 10^5 N / C. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery gen phy problem 16.32. field 745 N/C midway between two equal and opposite point charges separated by 16 cm. What is the magnitude of each charge? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: N/A univ phys confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * We make the following conceptual observations: At a point halfway between two opposite charges, a positive test charge will be repelled from one and attracted to the other. The repelling force on the test charge will be in the direction from the positive charge toward the negative charge, and the attracting force will also be toward the negative charge, so the two forces will reinforce one another. Thus the electric field at the halfway will be directed from the positive charge toward the negative, and will be double the field produced by either of the charges. The halfway point is 8 cm from each of the two charges. If the magnitude of the charge is q then the field contribution of each is k q / r^2, with r = 8 cm = .08 meters. Since both charges contribute equally to the field, with the fields produced by both charges being in the same direction (on any test charge at the midpoint one force is of repulsion and the other of attraction, and the charges are on opposite sides of the midpoint), the field of either charge has magnitude 1/2 (745 N/C) = 373 N/C. Thus the field of one of the charges is E = 373 N/C. Another expression for this field is E = k q / r^2. We solve for q to obtain q = E * r^2 / k. Substituting our values for k, E and r we obtain q = E * r^2 / k = 373 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2) = 373 N/C * .0064 m^2 / (9 * 10^9 N m^2 / C^2) = 2.6 * 10^-10 C, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: Electric field for one = k Q / r^2 Therefore the total electric field is 2 k Q / r^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** This calls for a symbolic expression in terms of the symbol Q. The field from either charge is k Q / r^2, directed toward the negative charge. The field of both charges together is therefore E_total = 2 k Q / r^2, where r=.08 meters. ** STUDENT COMMENT: That is a tough one. I will have to read up on this one. I guess you just added the 2 because they are two charges? INSTRUCTOR RESPONSE: There are two charges and you are asked for the field at their midpoint. We find the field due to each of the two charges, then we add the two fields. Had the charges been of the same sign, rather than equal and opposite, the two fields would have been equal, but opposite, and would therefore have added up to zero. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Magnitude electric field E = (kqQ/r^2)/ Q Therefore (kqQ/r^2) / Q = k q / r^2 E = 9 e^9 N m^2 / C^2 * 33.0 e-6 C / (.200 m)^2 E = 7.43 e6 N/C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: Flux = field . normal vector * area Area = .09 m^2 S1: flux = (-5 N/(Cm) xi + 3 N/(Cm) zk) . (-j) * .09 m^2 = 0 S2: flux = (-5 N/(Cm) xi + 3 N/(Cm) zk) . (k) * .09 m^2 = 3z N/(Cm) * .09 m^2 S3: flux = (-5 N/(Cm) xi + 3 N/(Cm) zk) . (j) * .09 m^2 = 0 s4: flux = (-5 N/(Cm) xi + 3 N/(Cm) zk) . (-k) * .09 m^2 = -3z N/(Cm) * .09 m^2 s5: flux = (-5 N/(Cm) xi + 3 N/(Cm) zk) . (i) * .09 m^2 = -5x N / (Cm) * .09 m^2 s6: flux = (-5 N/(Cm) xi + 3 N/(Cm) zk) . (-i) * .09 m^2 = 5x N / (Cm) * .09 m^2 Flux_tot = .027 Nm^2/C - .045 Nm^2/C Flux_tot = -.018 Nm^2/C FLux_tot = 4 pi k Q -.018 Nm^2/C = 4 pi k Q Q = -.018 Nm^2/C / (4 pi 10^9 Nm^2/C^2) Q = -1.6 e-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):ok ********************************************* 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: surface area of curved portion = 2 pi r L charged enclosed = 4 pi k L * alpha electric field = flux/area = 4 pi k L * alpha / (2 pi r L) = 2 k alpha / r inner surface of conductor needs -L * alpha when r>a and When r>b, charge enclosed = 2L * alpha electric field = 4 pi k * 2 L * alpha / (2 pi r L ) = 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 Therefore charge on outer cylinder = 2 alpha * L Charge density = 2 alpha 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):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: E = flux/area = 4 pi k lambda * L / (2 pi r L) = 2 k lambda /r We integrate this from a to b: = 2k lambda ln(r) evaulate from a to b: Vab = 2 k lambda (ln(b) - ln(a)) Vab = 2 k lambda ln (b/a) Substituting E: E = Vab / (ln(b/a)) * 1/r E = Vab / (r ln(b/a)) E = 20,000 V/m , r = 1.2 cm Vab = E * r * ln(b/a) Vab = 20,000 V/m *.012m * ln(1.8/.0145 cm) Vab = 1157 V confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The voltage V_ab is obtained by integrating the electric field from the radius of the central wire to the outer radius. From this we determine that E = Vab / ln(b/a) * 1/r, where a is the inner radius and b the outer radius. If E = 20,000 V/m at r = 1.2 cm then Vab = E * r * ln(b/a) = 20,000 V/m * ln(1.8 cm / .0145 cm) * .012 m = 1157 V. ** STUDENT QUESTION: Can you tell me what you integrated to get: E = Vab / ln(b/a) * 1/r ? INSTRUCTOR RESPONSE: Sure. The following assumes you know how to use Gaussian surfaces for axially symmetric charge distributions. If necessary see your text to fill in the details, but given the basic knowledge the explanation that follows is complete. I'll also be glad to clarify anything you wish to ask about: If the charge per unit length on the inner cylinder is lambda, then a coaxial cylinder of length L will contain charge Q = lamda * L. So the flux through the cylinder will be 4 pi k Q = 4 pi k lambda * L. Using symmetry arguments and assuming edge effects to be negligible, the electric field penetrates the curved surface of the cylinder at right angles. The area of the curved surface of such a coaxial cylinder of radius r is 2 pi r * L, so the electric field is field E = flux / area = 4 pi k lambda * L / (2 pi r L) = 2 k lambda / r. Integrating this field from inner radius a to outer radius b, we get the potential difference Vab: Our antiderivative function is 2 k lambda ln | r |, so the change in the antiderivative is Vab = 2 k lambda ( ln(b) - ln(a) ) = 2 k lambda ln(b / a). Thus Vab = 2 k lambda ln(b/a). This gives us 2 k lambda = Vab / (ln(b/a)), which will be used below. Since E = 2 k lambda / r, we substitute to get E = Vab / (ln(b/a)) * 1 / r, the expression about which you asked, and which we might wish to simplify into the form E = Vab / (r ln(b/a) ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!