#$&* course Phy 232 7/27 4:30 026. Query 27
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Given Solution: The electric field in the wire is equal to the voltage divided by the length of the wire. So a longer wire has a lesser electric field, which results in less acceleration of the free charges (in this case the electrons in the conduction band), and therefore a lower average charge velocity and less current. The greater the cross-sectional area the greater the volume of wire in any given length, so the greater the number of charge carriers (in this case electrons), and the more charges to respond to the electric field. This results in a greater current, in proportion to the cross-sectional area. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* 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: Current = number of charges / change in time Change in time = change in length / velocity Current = number of charges * velocity / change in length
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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 #$&* ok confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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 area gives greater number of charges. This causes a greater current, which implies 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: As stated in earlier questions, greater length causes lesser electrical field. Which causes less current, 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: 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: `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: 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: 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: 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 22.32 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: E = −Bi+Cj−Dk; A=L^2 face S1 : n = − j epsilonE =E*A=E* An= (−Bi+Cj−Dk) * −Aj = −CL^2 face S2 : n = +k epsilonE =E*A=E* An =( −Bi +Cj−Dk) * Ak = −DL^2 face S3 : n = + j epsilon E =E*A=E* An = (−Bi +Cj−Dk )* Aj = +CL^2 face S4 : n = −k epsilon E =E*A=E* An = (−Bi +Cj−Dk) * −Ak = +DL^2 face S5 : n = +i epsilonE =E*A=E*An= (−Bi+Cj−Dk) *Ai = −BL^2 face S6 :n = −i epsilonE =E*A=E* An= (−Bi+Cj−Dk)* −Ai = +BL^2 epsilonE = −CL^2 – DL^2 + CL^2 + DL^2 – BL^2 + BL^2 = 0 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: **** Advance correction of a common misconception: Flux is not a vector quantity so your flux values will not be multiples of the i, j and k vectors. The vectors normal to S1, S2, ..., S6 are respectively -j, k, j, -k, i and -i. For any face the flux is the dot product of the field with the normal vector, multiplied by the area. The area of each face is (.3 m)^2 = .09 m^2 So we have: For S1 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-j) * .09 m^2 = 0. For S2 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( k) * .09 m^2 = 3 z N / (C m) * .09 m^2. For S3 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( j) * .09 m^2 = 0. For S4 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-k) * .09 m^2 = -3 z N / (C m) * .09 m^2. For S5 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( i) * .09 m^2 = -5 x N / (C m) * .09 m^2. For S6 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-i) * .09 m^2 = 5 x N / (C m) * .09 m^2. On S2 and S4 we have z = .3 m and z = 0 m, respectively, giving us flux .027 N m^2 / C on S2 and flux 0 on S4. On S5 and S6 we have x = .3 m and x = 0 m, respectively, giving us flux -.045 N m^2 / C on S5 and flux 0 on S6. The total flux is therefore .027 N m^2 / C - .045 N m^2 / C = -.018 N m^2 / C. Since the total flux is 4 pi k Q, where Q is the charge enclosed by the surface, we have 4 pi k Q = -.018 N m^2 / C and Q = -.018 N m^2 / C / (4 pi k) = -.018 N m^2 / C / (4 pi * 9 * 10^9 N m^2 / C^2) = -1.6 * 10^-13 C, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `qquery univ 22.45 11th edition 22.37 (23.27 10th edition) Spherical conducting shell inner radius a outer b, concentric with larger conducting shell inner radius c outer d. Total charges +2q, +4q. Give your solution. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: (i) charge on inner surface of the small shell: Apply Gauss’s law to a spherical Gaussian surface with radius a < r < b. This surface lies within the conductor of the small shell, where E = 0, so epsilonE = 0. Thus by Gauss’s law Qencl = 0, so there is zero charge on the inner surface of the small shell. (ii) charge on outer surface of the small shell: The total charge on the small shell is +2q. We found in part (i) that there is zero charge on the inner surface of the shell, so all +2q must reside on the outer surface. (iii) charge on inner surface of large shell: Apply Gauss’s law to a spherical Gaussian surface with radius c
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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): ok ------------------------------------------------ Self-critique rating #$&* ok "