Engineering Mathematics: Engineering Electromagnetics-6th-edition- by william-h-hayt-john-a-buck: All Questions Chapter 1 to 10 and Solutions

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Given the vectors M ax ay a and N ax ay a, find: a a unit vector in the direction of M N. Given three points, A, , , B, , , and C, , : a Specify the vector A extending from the origin to the point A.... The vector from the origin to the point A is given as 6, , , and the unit vector directed from the origin toward point B is , , . If points A and B are ten units apart, find the coordinates of point .. given points A, , and B, , , find: d the coordinates of the point on the line connecting A to B at which the line intersects the plane . Note that the midpoint, , , , as determined from part c happens to have coordinate of . This is the point we are looking for. .. Use the definition of the dot product to find the interior angles at A and B of the triangle defined by the three points c the angle between RDA and RDC : Use RDA RAD , , and RDC , 6, . The angle is found through the dot product of the associated unit vectors, or The four vertices of a regular tetrahedron are located at .. Three vectors extending from the origin are given as , , . a a unit vector perpendicular to both r and r Evaluate D at the point where , ., and , expressing the result in cylindrical and cartesian coordinates: At the given point, and in cylindrical coordinates, D .a . To express this in cartesian, we use .. Express in cartesian components: .. A field is given in cylindrical coordinates as NOTE: The limits on the integration must be converted to radians as was done here, but not shown. b Find the total area of the enclosing surface: d Find the length of the longest straight line that lies entirely within the volume: This will be between the points A , , and B , , .. Performing point transformations to cartesian coordinates, these become Ax ., y ., and Bx ., y ., .. Taking A and B as vectors directed from the origin, the requested length Express ar in cartesian components at Determine the cartesian components of the vector from .. Express the unit vector ax in spherical components at the point: continued Express the unit vector ax in spherical components at the point: c the distance from A to C on a great circle path: Note that A and C share the same r and coordinates; thus moving from A to C involves only a change in of 6. The requested arc length is then .. Four nC positive charges are located in the plane at the corners of a square cm on a side. A fifth nC positive charge is located at a point cm distant from the other charges. Calculate the magnitude of the total force on this fifth charge for ǫ ǫ: Arrange the charges in the xy plane at lo√cations charge will be on the axis at location , which puts it at cm distance from the other four. By symmetry, the force on the fifth charge will be directed, and will be four times the component of force produced by each of the four other charges. .. A charge Q . C is located at the origin, while Q . C is at A., .6, . Find the locus of points in the plane at which the x component of the force on a third positive charge is ero. To solve this problem, the coordinate of the third charge is immaterial, so we can place it in the xy plane at coordinates x, y, . We take its magnitude to be Q. The vector directed from the first charge to the third is R xax yay ; the vector directed from the second charge to the third is ay . The force on the third charge is now .. Point charges of nC each are located at A, , , B , , , C, , , and D, , in free space. Find the total force on the charge at A. The force will be: where R Note, however, that all three charges must lie in a straight line, and the location of Q will be along the vector R extended past Q. The slope of this vector is 6 .. Therefore, we look for P at coordinates x, .x, . With this restriction, the force becomes: To obtain Ex , we require the expression in the large brackets to be ero. This expression simplifies to the following quadratic: b What single charge at the origin would provide the identical field strength? We require where R, R, R are the vectors to P from each of the charges in their original listed order. Specifically, .. A nC point charge is located at A, , in free space. a Find the locus of all points P x, y, at which Ex Vm: The total field at P will be: where R, the vector from the positive charge to point P is , y, , and R, the vector from the negative charge to point P , is , y, . The magnitudes of these vectors are R R y. Substituting these into the expression for EP produces .. A charge Q located at the origin in free space produces a field for which .. The volume charge density exists over all free space. Calculate the total charge present: This will be times the integral of v over the first octant, or .. A uniform volume charge density of . Cm note typo in book is present throughout the spherical shell extending from r cm to r cm. If v elsewhere: a find the total charge present throughout the shell: This will be b find r if half the total charge is located in the region cm r r: If the integral over r in part a is taken to r, we would obtain Following the same evaulation procedure as in part a, we obtain Q′ . mC. .. A spherical volume having a m radius contains a uniform volume charge density of Cm. a What total charge is enclosed in the spherical volume? This will be b Now assume that a large region contains one of these little spheres at every corner of a cubical grid mm on a side, and that there is no charge between spheres. What is the average volume charge density throughout this large region? Each cube will contain the equivalent of one little sphere. Neglecting the little sphere volume, the average density becomes b Find E at that point in the plane where the direction of E is given by ay a: With , the general field will be .. A uniform line charge is located on the axis. Find E in cartesian coordinates at P , , if the charge extends from a With the infinite line, we know that the field will have only a radial component in cylindrical coordinates or x and y components in cartesian. The field from an infinite line on the axis is generally E [l ǫ]a . Therefore, at point P : The student is invited to verify that when evaluating the above expression over the limits ∞ ∞, the component vanishes and the x and y components become those found in part a. .. Uniform line charges of nCm lie along the entire extent of the three coordinate axes. Assuming free space conditions, find E at P , , : Since all line charges are infinitelylong, we can write: .. Two identical uniform line charges with l nCm are located in free space at x , y . m. What force per unit length does each line charge exert on the other? The charges are parallel to the axis and are separated by . m. Thus the field from the charge at y . evaluated at the location of the charge at y . will be. The force on a differential length of the line at the positive y location is Thus the force per unit length acting on the line at postive y arising from the charge at negative y is The force on the line at negative y is of course the same, but with ay . .. A uniform surface charge density of nCm is present in the region x , y , and all . If ǫ ǫ, find E at: a PA, , : We use the superposition integral: Since the integration limits are symmetric about the origin, and since the y and components of the integrand exhibit odd parity change sign when crossing the origin, but otherwise symmetric, these will integrate to ero, leaving only the x component. This is evident just from the symmetry of the problem. Performing the integration first on the x component, we obtain using tables: The student is encouraged to verify that if the y limits were ∞ to ∞, the result would be that of the infinite charged plane, or Ex s ǫ. b PB , , : In this case, r ay , and symmetry indicates that only a y component will exist. The integral becomes .. Surface charge density is positioned in free space as follows: nCm at x , nCm at y , and nCm at . Find the magnitude of E at the three points, , , , , , , and , , . Since all three sheets are infinite, the field magnitude associated with each one will be s ǫ, which is positionindependent. For this reason, the net field magnitude will be the same everywhere, whereas the field direction will depend on which side of a given sheet one is positioned. We take the first point, for example, and find .. Find E at the origin if the following charge distributions are present in free space: point charge, nC at P , , 6; uniform line charge density, nCm at x , y ; uniform surface charge density, . nCm at x . The sum of the fields at the origin from each charge in order is: .6. A uniform line charge density of nCm is at y , m in free space, while nCm is located at y , m. A uniform surface charge density of . nCm is at y . m, and . nCm is at y . m. Find E at the origin: Since each pair consists of equal and opposite charges, the effect at the origin is to double the field produce by one of each type. Taking the sum of the fields at the origin from the surface and line charges, respectively, we find: .. Given the electric field intensity E yax xay Vm, find: a the equation of the streamline passing through the point A, , : Write: b the equation of the surface on which . Thus or we have a circularcylindrical surface, centered on the axis, and of radius . c A sketch of the part a equation would yield a parabola, centered at the origin, whose axis is the positive x axis, and for which the slopes of the asymptotes are . d A sketch of the trace produced by the intersection of the surface of part b with the plane would yield a circle centered at the origin, of radius . .. In cylindrical coordinates with E, E , a E , a , the differential equation describ ing the direction lines is E E dd in any constant plane. Derive the equation of the line passing through the point P , , in the field E cos a sin a : Using the given information, we write .. An empty metal paint can is placed on a marble table, the lid is removed, and both parts are discharged honorably by touching them to ground. An insulating nylon thread is glued to the center of the lid, and a penny, a nickel, and a dime are glued to the thread so that they are not touching each other. The penny is given a charge of nC, and the nickel and dime are discharged. The assembly is lowered into the can so that the coins hang clear of all walls, and the lid is secured. The outside of the can is again touched momentarily to ground. The device is carefully disassembled with insulating gloves and tools. a What charges are found on each of the five metallic pieces? All coins were insulated during the entire procedure, so they will retain their original charges: Penny: nC; nickel: ; dime: . The penny’s charge will have induced an equal and opposite negative charge nC on the inside wall of the can and lid. This left a charge layer of nC on the outside surface which was neutralied by the ground connection. Therefore, the can retained a net charge of nC after disassembly. b If the penny had been given a charge of nC, the dime a charge of nC, and the nickel a charge of nC, what would the final charge arrangement have been? Again, since the coins are insulated, they retain their original charges. The charge induced on the inside wall of the can and lid is equal to negative the sum of the coin charges, or nC. This is the charge that the canlid contraption retains after grounding and disassembly. .. A point charge of nC is located at the origin. four uniform line charges are located in the x plane as follows: nCm at y and m, nCm at y and m. a Find D at P , , : Note that this point lies in the center of a symmetric arrangement of line charges, whose fields will all cancel at that point. Thus D arise from the point charge alone, and will be × ay a ay . × a Cm .a pCm b How much electric flux crosses the plane y and in what direction? The plane intercepts all flux that enters the y halfspace, or exactly half the total flux of nC. The answer is thus 6 nC and in the ay direction. c How much electric flux leaves the surface of a sphere, m in radius, centered at C, , ? This sphere encloses the point charge, so its flux of nC is included. The line charge contributions are most easily found by translating the whole assembly sphere and line charges such that the sphere is centered at the origin, with line charges now at y and . The flux from the line charges will equal the total line charge that lies within the sphere. The length of each of the inner two line charges at y will be .. Let D xyax x ay ya Cm and evaluate surface integrals to find the total charge enclosed in the rectangular parallelepiped x , y , m: Of the 6 surfaces to consider, only will contribute to the net outward flux. Why? First consider the planes at y and . The y component of D will penetrate those surfaces, but will be inward at y and outward at y , while having the same magnitude in both cases. These fluxes will thus cancel. At the x plane, Dx and at the plane, D , so there will be no flux contributions from these surfaces. This leaves the remaining surfaces at x and . The net outward flux becomes: wo uniform line charges, each nCm, are located at y , m. Find the total flux leaving a sphere of radius m if it is centered at a A, , : The result will be the same if we move the sphere to the origin and the line charges to , , . The length of the line charge within the sphere is given by l sin[cos] .6. With two line charges, symmetrically arranged, the total charge enclosed is given by Q .6 nCm nC b B, , : In this case the result will be the same if we move the sphere to the origin and keep the charges where they were. Th√e length of the line joining the origin to the midpoint of the line charge in the y plane is l . The length of the line joining the origin to either endpoint of the line charge is then just the sp√here radius, or . The halfangle subtended at the origin by the line charge is then ψ cos√ . The length of each line charge in the sphere is then l√ × sin ψ . The total charge enclosed with two line charges is now Q′ nCm nC By using Gauss’s law, calculate the value of Dr on the surface r mm: The gaussian surface is a spherical shell of radius mm. The enclosed charge is the result of part a. We thus .. Let s Cm in the region where x and m, and let s elsewhere. Find D at P x, , , where x : The sheet charge can be thought of as an assembly of infinitelylong parallel strips that lie parallel to the y axis in the y plane, and where each is of thickness d. The field from each strip is that of an infinite line charge, and so we can construct the field at P from a single strip as: continued where r xax a and r′ ′a We distinguish between the fixed coordinate of P , , and the variable coordinate, ′, that determines the location of each charge strip. To find the net field at P , we sum the contributions of each strip by integrating over ′: The student is invited to verify that for very small x or for a very large sheet allowing ′ to approach infinity, the above expression reduces to the expected form, DP s . Note also that the expression is valid for all x positive or negative values. .. In cylindrical coordinates, let v for mm, v sin nCm for mm . mm, and v for . mm. Find D everywhere: Since the charge varies only with radius, and is in the form of a cylinder, symmetry tells us that the flux density will be radiallydirected and will be constant over a cylindrical surface of a fixed radius. Gauss’ law applied to such a surface of unit length in gives: a for mm, D within this range. , since no charge is enclosed by a cylindrical surface whose radius lies b for mm . mm, we have continued c for . mm, the gaussian cylinder now lies at radius outside the charge distribution, so the integral that evaluates the enclosed charge now includes the entire charge distribution. To accomplish this, we change the upper limit of the integral of part b from to . mm, finally obtaining: .. A nonuniform volume charge density, v r Cm, lies within the spherical surface r m, and v everywhere else. a Find Dr everywhere. For r m, we apply Gauss’ law to a spherical surface of radius r within this range to find Thus Dr r for r m. For r m, the gaussian surface lies outside the charge distribution. The set up is the same, except the upper limit of the above integral is instead of r. This results in Dr r for r m. b What surface charge density, s, should be on the surface r such that Dr,r Dr,r? At r , we have Dr,r , from part a. The flux density in the region r arising from a surface charge at r is found from Gauss’ law through The total flux density in the region r arising from the two distributions is .. Spherical surfaces at r , , and 6 m carry uniform surface charge densities of nCm, nCm, and s, respectively. a Find D at r , and m: Noting that the charges are sphericallysymmetric, we ascertain that D will be radiallydirected and will vary only with radius. Thus, we apply Gauss’ law to spherical shells in the following regions: r : Here, no charge is enclosed, and so D b Determine s such that D at r m. Since fields will decrease as r, the question could be rephrased to ask for s such that D at all points where r 6 m. In this region, the total field will be v nCm for mm and no other charges are present: a find D for mm: Applying Gauss’ law to a cylindrical surface of unit length in , and of radius mm, we find c Evaluate D at . mm, .6 mm, and . mm: At . mm, no charge is enclosed by a cylindrical gaussian surface of that radius, so D .mm . At .6 mm, we evaluate the part b result at .6 to obtain: .6. Given the electric flux density, D xy ax x ay 6 a Cm: a use Gauss’ law to evaluate the total charge enclosed in the volume x, y, a: We call the surfaces at x a and x the front and back surfaces respectively, those at y a and y the right and left surfaces, and those at a and the top and bottom surfaces. To evaluate the total charge, we integrate D • n over all six surfaces and sum the results continued Noting that the back and bottom integrals are ero, and that the left and right integrals cancel, we evaluate the remaining two front and top to obtain Q 6a a. b use Eq. to find an approximate value for the above charge. Evaluate the derivatives at P a, a, a: In this application, Eq. states that Q . • D Ov. We find • D x , which when evaluated at P becomes • D a .a . Thus Q a .a a .a a c Show that the results of parts a and b agree in the limit as a → . In this limit, both expressions reduce to Q a, and so they agree. .. A cube is defined by x, y, .. If D xyax a apply Gauss’ law to find the total flux leaving the closed surface of the cube. We call the surfaces at x . and x the front and back surfaces respectively, those at y . and y the right and left surfaces, and those at . and the top and bottom surfaces. To evaluate the total charge, we integrate D n over all six surfaces and sum the results. We note that there is no component of D, so there will be no outward flux contributions from the top and bottom surfaces. The fluxes through the remaining four are .. Let a vector field by given by G xy ay . Evaluate both sides of Eq. for this G field and the volume defined by x and ., y and ., and and .. Evaluate the partial derivatives at the center of the volume. First find The center of the cube is located at .,.,., and the volume is Ov . .. Eq. then .. A spherical surface of radius mm is centered at P , , in free space. Let D xax Cm. Use the results of Sec. . to estimate the net electric flux leaving the spherical surface: We use . .. A cube of volume a has its faces parallel to the cartesian coordinate surfaces. It is centered at P , , . Given the field D xax Cm: a calculate div D at P : In the present case, this will be b evaluate the fraction in the rightmost side of Eq. for a m, . m, and mm: With the field having only an x component, flux will pentrate only the two surfaces at x a, each of which has surface area a. The cube volume is Ov a. The equation reads: a Find div D: Using the divergence formula for cylindrical coordinates see problem ., we find • D sin . b Find the volume charge density at P .6, , 6.: Since v • D, we evaluate the result of part a at this point to find vP sin . Cm. c How much charge is located inside the region defined by A point charge Q lies at the origin. Show that div D is ero everywhere except at the origin. For a point charge at the origin we know that D Qr ar . Using the formula for divergence in spherical coordinates see problem . solution, we find in this case that The above is true provided r . When r , we have a singularity in D, so its divergence is not defined. b Replace the point charge with a uniform volume charge density v for r a. Relate v to Q and a so that the total charge is the same. Find div D everywhere: To achieve the same net charge, we require that av Q, so v Qa Cm. Gauss’ law tells us that inside the charged sphere . How much electric flux leaves the closed surface , , . .? We note that D has only a radial component, and so flux would leave only through the cylinder sides. Also, D does not vary with or , so the flux is found by a simple product of the side area and the flux density. We further note that D at , so only the outer side at will contribute. We use the result of part b, and write the flux as How much charge is contained within the volume used in part c? By Gauss’ law, this will be the same as the net outward flux through that volume, or again, C. .. Within the spherical shell, r m, the electric flux density is given as which we evaluate at r to find v r . Cm. b What is the electric flux density at r ? Substitute r into the given expression to find D ar Cm c How much electric flux leaves the sphere r ? Using the result of part b, this will be C d How much charge is contained within the sphere, r ? From Gauss’ law, this will be the same as the outward flux, or again, Q C. .6. Given the field 6d. the total electric flux leaving the surface r Since the total enclosed charge is ero from part b, the net outward flux is also ero, from Gauss’ law. .. Let D .rar mCm for r . m and D . ar r Cm for r . m note error in problem statement. a Find v for r .6 m: This radius lies within the first region, and so b Find v for r . m: This is in the region where the second field expression is valid. The r dependence of this field yields a ero divergence shown in Problem ., and so the volume charge density is ero at . m. c What surface charge density could be located at r . m to cause D for r . m? The total surface charge should be equal and opposite to the total volume charge. The latter is .. The electric flux density is given as D a Cm for m, and k a for m. a Find k so that D is continuous at m: We require b Find and sketch v as a function of : In cylindrical coordinates, with only a radial component of D, we use . Evaluate the surface integral side for the corresponding closed surface: We call the surfaces at x and x the front and back surfaces respectively, those at y and y the right and left surfaces, and those at and the top and bottom surfaces. To evaluate the surface integral side, we integrate D n over all six surfaces and sum the results. Note that since the x component of D does not vary with x, the outward fluxes from the front and back surfaces will cancel each other. The same is true for the left and right surfaces, since Dy does not vary with y. This leaves only the top and bottom surfaces, where the fluxes are: .. If D sin a cos a Cm, evaluate both sides of the divergence theorem for the region m, rad, m: Taking the surface integral side first, the six sides over which the flux must be evaluated are only four, since there is no component of D. We are left with the sides at and rad left and right sides, respectively, and those at and back and front sides. We evaluate use two different methods to find the total charge within the region r m, rad, rad: We use the divergence theorem and first evaluate the surface integral side. We are evaluating the net outward flux through a curvilinear “cube”, whose boundaries are defined by the specified ranges. The flux contributions will be only through the surfaces of constant , however, since D has only a component. On a constanttheta surface, the differential area is da r sin drd, where is fixed at the surface location. Our flux integral becomes .. The value of E at P , , is given as E a a a Vm. Determine the incremental work required to move a C charge a distance of 6 m: a in the direction of a : The incremental work is given by dW q E • dL, where in this case, dL d a 6 × 6 a . Thus .. Let E ax ay a in the neighborhood of point P 6, , . Find the incremental work done in moving a C charge a distance of mm in the direction specified by: a ax ay a: We write b Q, , toward P , , : A little thought is in order here: Note that the field has only a ra√dial component and does not depend on or . Note also that P and Q are at the same radius from the axis, but have different and coordinates. We could just as well position the two points at the same location and the problem would not change. If this were so, then moving along a st√raight line between P and Q would thus involve moving along a chord of a circle whose radius is . Halfway along this line is a point of symmetry in the field make a sketch to see this. This means that when starting from either point, the initial force will be the same. Thus the answer is dW . J as in part a. This is also found by going through the same procedure as in part a, but with the direction roles of P and Q reversed. .. A point charge Q is located at the origin in free space. Find the work done in carrying a charge Q from: a BrB , B , B to CrA, B , B with and held constant; b CrA, B , B to DrA, A, B with r and held constant; c DrA, A, B to ArA, A, A with r and held constant: The general expression for the work done in this instance is The answers to parts b and c involving paths over which r is held constant are both . .. A uniform surface charge density of nCm is present on the spherical surface r .6 cm in free space. a Find the absolute potential at P r cm, , : Since the charge density is uniform and is sphericallysymmetric, the angular coordinates do not matter. The potential function for r .6 cm will be that of a point charge of Q as , or .. Given a surface charge density of nCm on the plane x , a line charge density of nCm on the line x , y , and a C point charge at P , , , find VAB for points A, , and B, , : We need to find a potential function for the combined charges. That for the point charge we know to be Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have The terms in this expression are not referenced to a common origin, since the charges are at different positions. The parameters r, , and x are scalar distances from the charges, and will be treated as such here. For point A we have r .. Let a uniform surface charge density of nCm be present at the plane, a uniform line charge density of nCm be located at x , , and a point charge of C be present at P , , . If V at M, , , find V at N, , : We need to find a potential function for the combined charges which is ero at M. That for the point charge we know to be Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have The total potential function will be the sum of the three. Combining the integration constants, we obtain continued The terms in this expression are not referenced to a common origin, since the charges are at different positions. The parameters r, , and are scalar distances from the charges, and will be treate√d as such he√re. To evaluate the constant, C, we first look at point M, where VT . At M, r , , and . We thus have .. Three point charges, . C each, are located at , , , , , , and , , , in free space. a Find an expression for the absolute potential as a function of along the line x , y : From a point located at position along the given line, the distances to the three charges are R , , R √ , and R , . The total potential will be .. Three identical point charges of pC each are located at the corners of an equilateral triangle . mm on a side in free space. How much work must be done to move one charge to a point equidistant from the other two and on the line joining them? This will be the magnitude of the charge times the potential difference between the finishing and starting positions, or .6. Uniform surface charge densities of 6, , and nCm are present at r , , and 6 cm, respectively, in free space. a Assume V at infinity, and find Vr. We keep in mind the definition of absolute potential as the work done in moving a unit positive charge from infinity to location r. At radii outside all three spheres, the potential will be the same as that of a point charge at the origin, whose charge is the sum of the three sphere charges: .. Fig. . shows three separate charge distributions in the plane in free space. a find the total charge for each distribution: Line charge along the y axis: c the total charge lying within the surface r .6: The easiest way is to use Gauss’ law, and integrate the flux density over the spherical surface r .6. Since the field is constant at constant radius, we obtain the product: the total charge lying within the closed surface .6, : The easiest way to do this calculation is to evaluate D at .6 noting that it is constant, and then multiply by the cylinder area: Using part a, we have D ..6.. Given the potential field V r cos and a point P ., , 6 in free space, find at P : a V : Substitute the coordinates into the function and find VP . cos V. b E: .6. A dipole having Qdǫ V • m is located at the origin in free space and aligned so that its moment is in the a direction. a Sketch Vr ,, versus on polar graph paper homemade if you wish. b Sketch Er ,, versus on polar graph paper: Taking the magnitude of the above, we find EP . Vm. c Now treat the two√charges as a dipole at the origin and find V at P : In spherical coordinates, P is located at r we have .. A dipole located at the origin in free space has a moment p × a C • m. At what points on the line y , x is: The y and values are thus y .√ 6.m b Er mVm? From the above field expression, the radial component magnitude is twice that of the theta component. Using the same development, we then find .. A dipole having a moment p ax ay a nC m is located at Q, , in free space. Find V at P , , : We use the general expression for the potential in the far field: .. A dipole, having a moment p a nC m is located at the origin in free space. Give the magnitude of E and its direction aE in cartesian components at r m, , and : a ; b ; c . Begin with Show that the surface .6 is an equipotential surface: This will be the surface of a cone, centered at the origin, along which E, in the ar direction, will exist. Therefore, the given surface cannot be an equipotential the problem was illconceived. Only a surface of constant r could be an equipotential in this field. c Find VAB .. A copper sphere of radius cm carries a uniformlydistributed total charge of C in free space. a Use Gauss’ law to find D external to the sphere: with a spherical Gaussian surface at radius r, D will be the total charge divided by the area of this sphere, and will be ar directed. Thus .. Given the potential field in free space, V V note that aphi should not be present, find: a the energy stored in the region cm, ., m: First we find .c. the maximum value of the energy density in the specified region: The energy density is This will maximie at the lowest value of in the specified range, which is cm. So .. Four . nC point charges are located in free space at the corners of a square cm on a side. a Find the total potential energy stored: This will be given by where Vn in this case is the potential at the location of any one of the point charges that arises from the other three. This will be for charge Taking the summation produces a factor of , since the situation is the same at all four points. Consequently, b A fifth . C charge is installed at the center of the square. Again find the total stored energy: This will be the energy found in part a plus the amount of work done in moving the fifth charge into position from infinity. The latter is just the potential at the square center arising from the original four charges, times the new charge value, or .. Given the current density J [sinxey ax cosxey ay ] kAm: a Find the total current crossing the plane y in the ay direction in the region x , : This is found through b Find the total current leaving the region x, x , by integrating J dS over the surface of the cube: Note first that current through the top and bottom surfaces will not exist, since J has no component. Also note that there will be no current through the x plane, since Jx there. Current will pass through the three remaining surfaces, and will be found through c Repeat part b, but use the divergence theorem: We find the net outward current through the surface of the cube by integrating the divergence of J over the cube volume. We have Make the assumption that the electrons are emitted continuously as a beam with a . mm radius and a total current of 6 A. Find J and : negative since we have electrons flowing in the positive direction Next we use J v v, d Show that the divergence theorem is satisfied for J and the surface specified in part b. In part c, the net outward flux was found to be ero, and in part b, the divergence of J was found to be ero as will be its volume integral. Therefore, the divergence theorem is satisfied. .. Assuming that there is no transformation of mass to energy or viceversa, it is possible to write a continuity equation for mass. a If we use the continuity equation for charge as our model, what quantities correspond to J and v ? These would be, respectively, mass flux density in kgm s and mass density in kgm. b Given a cube cm on a side, experimental data show that the rates at which mass is leaving each of the six faces are ., ., ., ., ., and . mgs. If we assume that the cube is an incremental volume element, determine an approximate value for the time rate of change of density at its center. We may write the continuity equation for mass as follows, also invoking the divergence theorem: .. The continuity equation for mass equates the divergence of the mass rate of flow mass per second per square meter to the negative of the density mass per cubic meter. After setting up a cartesian coordinate system inside a star, Captain Kirk and his intrepid crew make measurements over the faces of a cube centered at the origin with edges km long and parallel to the coordinate axes. They find the mass rate of flow of material outward across the six faces to be , , , 6, , and kgm s. a Estimate the divergence of the mass rate of flow at the origin: We make the estimate using the definition of divergence, but without taking the limit as the volume shrinks to ero: Using data tabulated in Appendix C, calculate the required diameter for a m long nichrome wire that will dissipate an average power of W when V rms at 6 H is applied to it: The required resistance will be .. A steel wire has a radius of mm and a conductivity of 6 Sm. The steel wire has an aluminum σ . Sm coating of mm thickness. Let the total current carried by this hybrid conductor be A dc. Find: a Jst . We begin with the fact that electric field must be the same in the aluminum and steel regions. This comes from the requirement that E tangent to the boundary between two media must be continuous, and from the fact that when integrating E over the wire length, the applied voltage value must be obtained, regardless of the medium within which this integral is evaluated. We can therefore write The net current is now expressed as the sum of the currents in each region, written as the sum of the products of the current densities in each region times the appropriate crosssectional area: .. Two perfectlyconducting cylindrical surfaces are located at and cm. The total current passing radially outward through the medium between the cylinders is A dc. Assume the cylinders are both of length l. a Find the voltage and resistance between the cylinders, and E in the region between the cylinders, if a conducting material having σ . Sm is present for cm: Given the current, and knowing that it is radiallydirected, we find the current density by dividing it by the area of a cylinder of radius and length l: .. The spherical surfaces r and r cm are perfectly conducting, and the total current passing radially outward through the medium between the surfaces is A dc. a Find the voltage and resistance between the spheres, and E in the region between them, if a conducting material having σ . Sm is present for r cm. We first find J as a function of radius by dividing the current by the area of a sphere of radius r: .. A hollow cylindrical tube with a rectangular crosssection has external dimensions of . in by in and a wall thickness of . in. Assume that the material is brass, for which σ . Sm. A current of A dc is flowing down the tube. a What voltage drop is present across a m length of the tube? Converting all measurements to meters, the tube resistance over a m length will be: The voltage drop is now V I R . × . V. b Find the voltage drop if the interior of the tube is filled with a conducting material for which σ . × Sm: The resistance of the filling will be: b Find a unit vector directed outward to the surface, assuming the origin is inside the surface: Such a unit normal can be construced from the result of part c: b Show that the surface is an equipotential surface: There are two reasons for this: E at is everywhere directed, and so moving a charge around on the surface involves doing no work; When evaluating the given potential function at , the result is for all x and y. c Assume that the surface is a conductor and find the total charge on that portion of the conductor defined by x , y : We have where the integration “constants” are functions of all variables other than the integration variable. The general procedure is to adjust the functions, f , such that the result for V is the same in all three integrations. In this case we see that f x, y f x, f y, accomplishes this, and the potential function is V xy as given. d Determine VPQ, given Q, , : Using the potential function of part c, we have a Determine the equations of the equipotential surfaces on which V and 6 V: Setting the given potential function equal to and 6 and simplifying results in: b Assume these are conducting surfaces and find the surface charge density at that point on the V 6 V surface where x and . It is known that V 6 V is the fieldcontaining region: First, on the 6 V surface, we have .. The mobilities for intrinsic silicon at a certain temperature are e . mV s and h . mV s. The concentration of both holes and electrons is . 6 m. Determine both the conductivity and the resistivity of this silicon .. Electron and hole concentrations increase with temperature. For pure silicon, suitable expressions are h e 6T .eT Cm. The functional dependence of the mobilities on temperature is given by h . T . mV s and e . T . mV s, where the temperature, T , is in degrees Kelvin. The conductivity will thus be .6. A little donor impurity, such as arsenic, is added to pure silicon so that the electron concentration is conduction electrons per cubic meter while the number of holes per cubic meter is only . . If e . mV s for this sample, and h . mV s, determine the conductivity and resistivity .. Atomic hydrogen contains . atomsm at a certain temperature and pressure. When an electric field of kVm is applied, each dipole formed by the electron and positive nucleus has an effective length of . m. a Find P: With all identical dipoles, we have This normal will point in the direction of increasing f , which will be away from the origin, or into region you can visualie a portion of the surface as a tri√angle whose vertices are on the three coordinate axes at x , y , and .. So EN 6[ ] . Vm. Since the magnitude is negative, the normal component points into region from the surface. Then .. Let the spherical surfaces r cm and r cm be separated by two perfect dielectric shells, ǫR for r 6 cm and ǫR for 6 r cm. If E rar Vm, find: a E: Since E is normal to the interface between ǫR and ǫR, D will be continuous across the boundary, and so .. Capacitors tend to be more expensive as their capacitance and maximum voltage, Vmax , increase. The voltage Vmax is limited by the field strength at which the dielectric breaks down, EBD . Which of these dielectrics will give the largest CVmax product for equal plate areas: a air: ǫR , EBD MVm; b barium titanate: ǫR , EBD MVm; c silicon dioxide: ǫR ., EBD 6 MVm; d polyethylene: ǫR .6, EBD . MVm? Note that Vmax EBDd, where d is the plate separation. Also, C ǫR ǫAd, and so Vmax C ǫR ǫAEBD , where A is the plate area. The maximum CVmax product is found through the maximum ǫR EBD product. Trying this with the given materials yields the winner, which is barium titanate. .. A parallel plate capacitor is filled with a nonuniform dielectric characteried by ǫR × 6x, where x is the distance from one plate. If S . m, and d mm, find C: Start by assuming charge density s on the top plate. D will, as usual, be xdirected, originating at the top plate and terminating on the bottom plate. The key here is that D will be constant over the distance between plates. This can be understood by considering the xvarying dielectric as constructed of many thin layers, each having constant permittivity. The permittivity changes from layer to layer to approximate the given function of x. The approximation becomes exact as the layer thicknesses approach ero. We know that D, which is normal to the layers, will be continuous across each boundary, and so D is constant over the plate separation distance, and will be given in magnitude by s . The electric field magnitude is now a. The width of the region containing ǫR in Fig. . is . m. Find ǫR if ǫR . and the total capacitance is 6 nF: The plate areas associated with each capacitor are A . . m and A . .6 m. Having parallel capacitors, the capacitances will add, so b Find the width of each region containing ǫR and ǫR if Ctotal nF, ǫR ǫR, and C C: Let w be the width of region . The above conditions enable us to write .. Let ǫR . for y mm, ǫR for y mm, and ǫR for y mm. Conducting surfaces are present at y and y mm. Calculate the capacitance per square meter of surface area if: a ǫR is that of air; b ǫR ǫR; c ǫR ǫR; d ǫR is silver: The combination will be three capacitors in series, for which continued The charge on a unit length of the inner conductor is Q .s . The capacitance is now Q .s C ǫ pFm V .s ǫ Note that throughout this problem, I left all dimensions in cm, knowing that all cm units would cancel, leaving the units of capacitance to be those used for ǫ. .. Two coaxial conducting cylinders of radius cm and cm have a length of m. The region between the cylinders contains a layer of dielectric from c to d with ǫR . Find the capacitance if a c cm, d cm: This is two capacitors in series, and so .. Conducting cylinders lie at and mm; both extend from to m. Perfect dielectrics occupy the interior region: ǫR for 6 mm, ǫR for 6 mm, and ǫR for mm. a Calculate C: First we know that D s a Cm, with expressed in mm. Then, with in mm, b A portion of the dielectric is now removed so that ǫR ., , and ǫR , . Again, find C: We recognie here that removing that portion leaves us with two capacitors in parallel whose C’s will add. We use the fact that with the dielectric completely removed, the capacitance would be CǫR . 6.6 pF. With onefourth the dielectric removed, the total capacitance will be .. A cm diameter conductor is suspended in air with its axis cm from a conducting plane. Let the potential of the cylinder be V and that of the plane be V. Find the surface charge density on the: a cylinder at a point nearest the plane: The cylinder will image across the plane, producing an equivalent twocylinder problem, with the second one at location cm below the plane. We will 6. Construct a curvilinear square map for a coaxial capacitor of cm inner radius and cm outer radius. These dimensions are suitable for the drawing. a Use your sketch to calculate the capacitance per meter length, assuming ǫR : The sketch is shown below. Note that only a sector was drawn, since this would then be duplicated times around the circumference to complete the drawing. The capacitance is thus 6. Construct a curvilinearsquare map of the potential field about two parallel circular cylinders, each of . cm radius, separated by a centertocenter distance of cm. These dimensions are suitable for the actual sketch if symmetry is considered. As a check, compute the capacitance per meter both from your sketch and from the exact formula. Assume ǫR . Symmetry allows us to plot the field lines and equipotentials over just the first quadrant, as is done in the sketch below shown to onehalf scale. The capacitance is found from the formula C NQNV ǫ, where NQ is twice the number of squares around the perimeter of the halfcircle and NV is twice the number of squares between the halfcircle and the left vertical plane. The result is 6.. Construct a curvilinear square map of the potential field between two parallel circular cylinders, one of cm radius inside one of cm radius. The two axes are displaced by . cm. These dimensions are suitable for the drawing. As a check on the accuracy, compute the capacitance per meter from the sketch and from the exact expression: The drawing is shown below. Use of the exact expression above yields a capacitance value of C .ǫ Fm. Use of the drawing produces: 6.. A solid conducting cylinder of cm radius is centered within a rectangular conducting cylinder with a cm by cm crosssection. a Make a fullsie sketch of one quadrant of this configuration and construct a curvilinearsquare map for its interior: The result below could still be improved a little, but is nevertheless sufficient for a reasonable capacitance estimate. Note that the fivesided region in the upper right corner has been partially subdivided dashed line in anticipation of how it would look when the nextlevel subdivision is done doubling the number of field lines and equipotentials. b Assume ǫ ǫ and estimate C per meter length: In this case NQ is the number of squares around the full perimeter of the circular conductor, or four times the number of squares shown in the drawing. NV is the number of squares between the circle and the rectangle, or . The capacitance is estimated to be 6.. The inner conductor of the transmission line shown in Fig. 6. has a square crosssection a a, while the outer square is a a. The axes are displaced as shown. a Construct a goodsied drawing of the transmission line, say with a . cm, and then prepare a curvilinearsquare plot of the electrostatic field between the conductors. b Use the map to calculate the capacitance per meter length if ǫ .6ǫ. c How would your result to part b change if a .6 cm? a The plot is shown below. Some improvement is possible, depending on how much time one wishes to spend. 6.6. Let the inner conductor of the transmission line shown in Fig. 6. be at a potential of V, while the outer is at ero potential. Construct a grid, .a on a side, and use iteration to find V at a point that is a units above the upper right corner of the inner conductor. Work to the nearest volt: The drawing is shown below, and we identify the requested voltage as V. 6.. Use the iteration method to estimate the potentials at points x and y in the triangular trough of Fig. 6.. Work only to the nearest volt: The result is shown below. The mirror image of the values shown occur at the points on the other side of the line of symmetry dashed line. Note that Vx V and Vy 6 V. 6.. Use iteration methods to estimate the potential at point x in the trough shown in Fig. 6.. Working to the nearest volt is sufficient. The result is shown below, where we identify the voltage at x to be V. Note that the potentials in the gaps are V. 6.. Using the grid indicated in Fig. 6., work to the nearest volt to estimate the potential at point A: The voltages at the grid points are shown below, where VA is found to be V. Half the figure is drawn since mirror images of all values occur across the line of symmetry dashed line. 6.. Conductors having boundaries that are curved or skewed usually do not permit every grid point to coincide with the actual boundary. Figure 6.6a illustrates the situation where the potential at V is to be estimated in terms of V, V, V, and V, and the unequal distances h, h, h, and h. a Show that 6.. Consider the configuration of conductors and potentials shown in Fig. 6.. Using the method described in Problem , write an expression for Vx not V: The result is shown below, where Vx V. a After estimating potentials for the configuation of Fig. 6., use the iteration method with a square grid cm on a side to find better estimates at the seven grid points. Work to the nearest volt: Construct a . cm grid, establish new rough estimates, and then use the iteration method on the . cm grid. Again, work to the nearest volt: The result is shown below, with values for the original grid points underlined: Use the computer to obtain values for a . cm grid. Work to the nearest . V: Values for the left half of the configuration are shown in the table below. Values along the vertical line of symmetry are included, and the original grid values are underlined. 6.. Perfectlyconducting concentric spheres have radii of and 6 cm. The region r cm is filled with a solid conducting material for which σ Sm, while the portion for which r 6 cm has σ Sm. The inner sphere is held at V while the outer is at V . a. Find E and J everywhere: From symmetry, E and J will be radiallydirected, and we note the fact that the current, I , must be constant at any crosssection; i.e., through any spherical surface at radius r between the spheres. Thus we require that in both regions, 6.. The crosssection of the transmission line shown in Fig. 6. is drawn on a sheet of conducting paper with metallic paint. The sheet resistance is Ksq and the dimension a is cm. a Assuming a result for Prob. 6b of pFm, what total resistance would be measured between the metallic conductors drawn on the conducting paper? We assume a paper thickness of t m, so that the capacitance is C t pF, and the surface resistance is Rs σ t Ksq. We now use 6.. two concentric annular rings are painted on a sheet of conducting paper with a highly conducting metal paint. The four radii are , ., ., and . cm. Connections made to the two rings show a resistance of ohms between them. a What is Rs for the conducting paper? Using the two radii . and . cm at which the rings are at their closest separation, we first evaluate the capacitance: b If the conductivity of the material used as the surface of the paper is Sm, what is the thickness of the coating? We use 6.6. The square washer shown in Fig. 6. is . mm thick and has outer dimensions of . . cm and inner dimensions of . . cm. The inside and outside surfaces are perfectlyconducting. If the material has a conductivity of 6 Sm, estimate the resistance offered between the inner and outer surfaces shown shaded in Fig. 6.. A few curvilinear squares are suggested: First we find the surface resistance, Rs σ t 6 × . × 6. Ksq. Having found this, we can construct the total resistance by using the fundamental square as a building block. Specifically, R Rs Nl Nw where Nl is the number of squares between the inner and outer surfaces and Nw is the number of squares around the perimeter of the washer. These numbers are found from the curvilinear square plot shown below, which covers oneeighth the washer. The resistance is thus R . 6.[ × ] . 6. K. 6.. A twowire transmission line consists of two parallel perfectlyconducting cylinders, each having a radius of . mm, separated by centertocenter distance of mm. The medium surrounding the wires has ǫR and σ . mSm. A V battery is connected between the wires. Calculate: a the magnitude of the charge per meter length on each wire: Use 6.. A coaxial transmission line is modelled by the use of a rubber sheet having horiontal dimensions that are times those of the actual line. Let the radial coordinate of the model be m. For the line itself, let the radial dimension be designated by as usual; also, let a .6 mm and b . mm. The model is cm in height at the inner conductor and ero at the outer. If the potential of the inner conductor is V: a Find the expression for V : Assuming charge density s on the inner conductor, we use Gauss’ Law to find D as , from which E Dǫ as ǫ in the radial direction. The potential difference between inner and outer conductors is Now, as a function of radius, and assuming ero potential on the outer conductor, the potential function will be: b Write the model height as a function of m not : We use the part a result, since the gravitational function must be the same as that for the electric potential. We replace V by the maximum height, and multiply all dimensions by to obtain: Conciliate your results with the uniqueness theorem: Uniqueness specifies that there is only one potential that will satisfy all the given boundary conditions. While both potentials have the same value at r , they do not as r → ∞. So they apply to different situations. .. Conducting planes at cm and cm are held at potentials of V and V, respectively. The region between the plates is filled with a perfect dielectric with ǫ ǫ. Find and sketch: a V : We begin with the general solution of the onedimensional Laplace equation in rectangular coordinates: V A B. Applying the boundary conditions, we write A B and A B. Subtracting the former equation from the latter, we find 6A or A Vcm. Using this we find B V. Finally, V V in cm or V V in m. .. The conducting planes x y and x y are at potentials of V and , respectively. Let ǫ ǫ and find: a V at P , , 6: The planes are parallel, and so we expect variation in potential in the direction normal to them. Using the two boundary condtions, our general potential function can be written: .. Conducting cylinders at cm and cm in free space are held at potentials of 6mV and mV, respectively. a Find V : Working in volts and meters, we write the general onedimensional solution to the Laplace equation in cylindrical coordinates, assuming radial variation: V A ln B. Applying the given boundary conditions, this becomes V cm .6 A ln. B and V cm . A ln. B. Subtracting the former equation from the latter, we find . A ln.. A ln ⇒ A .6. B is then found through either equation .. Coaxial conducting cylinders are located at . cm and . cm. The region between the cylinders is filled with a homogeneous perfect dielectric. If the inner cylinder is at V and the outer at V, find: a the location of the V equipotential surface: From Eq. 6 we have .. Two semiinfinite planes are located at α and α, where α . A narrow insulating strip separates them along the axis. The potential at α is V, while V at α. a Find V in terms of α and V: We use the onedimensional solution form for Laplace’s equation assuming variation along : V A B. The boundary conditions are then substituted: V Aα B and Aα B. Subtract the latter equation from the former to obtain: V Aα ⇒ A Vα. Then Vαα B ⇒ B V. Finally V V . Σ V .. The two conducting planes illustrated in Fig. . are defined by . . m, . m, . and . rad. The medium surrounding the planes is air. For region , . ., neglect fringing and find: a V : The general solution to Laplace’s equation will be V C C, and so Solve Laplace’s equation for the potential field in the homogeneous region between two concentric conducting spheres with radii a and b, b a, if V at r b and V V at r a. With radial variation only, we have Note that in the last integration step, I dropped the minus sign that would have otherwise occurred in front of A, since we can choose A as we wish. Next, apply the boundary conditions: . Find the capacitance between them: Assume permittivity ǫ. First, the electric field will be .. Concentric conducting spheres are located at r mm and r mm. The region between the spheres is filled with a perfect dielectric. If the inner sphere is at V and the outer sphere at V: a Find the location of the V equipotential surface: Solving Laplace’s equation gives c Find ǫR if the surface charge density on the inner sphere is Cm: s will be equal in magnitude to the electric flux density at r a. So s .6 × VmǫR ǫ Cm. Thus ǫR ! obviously a bad choice of numbers here – possibly a misprint. A more reasonable charge on the inner sphere would have been Cm, leading to ǫR .. .. Concentric conducting spheres have radii of and cm. There is a perfect dielectric for which ǫR between them. The potential of the inner sphere is V and that of the outer is V. Find: a V r: We use the general expression derived in Problem .6: V r Ar B. At the inner sphere, A. B, and at the outer sphere, A. B. Subtracting the latter equation from the former gives .. Two coaxial conducting cones have their vertices at the origin and the axis as their axis. Cone A has the point A, , on its surface, while cone B has the point B, , on its surface. Let VA V and VB V. Find: .. A potential field in free space is given as V ln tan V. a Find the maximum value of E on the surface for . r . m, 6 . First .. A rectangular trough is formed by four conducting planes located at x and cm and y and cm in air. The surface at y cm is at a potential of V, the other three are at ero potential, and the necessary gaps are placed at two corners. Find the potential at x cm, y cm: This situation is the same as that of Fig. .6, except the nonero boundary potential appears on the top surface, rather than the right side. The solution is found from Eq. by simply interchanging x and y, and b and d, obtaining: Additional accuracy is found by including more terms in the expansion. Using thirteen terms, and using six significant figure accuracy, the result becomes V , . . V. The series converges rapidly enough so that terms after the sixth one produce no change in the third digit. Thus, quoting three significant figures, . V requires six terms, with subsequent terms having no effect. .. The four sides of a square trough are held at potentials of , , , and 6 V; the highest and lowest potentials are on opposite sides. Find the potential at the center of the trough: Here we can make good use of symmetry. The solution for a single potential on the right side, for example, with all other sides at V is given by Eq. : In the current problem, we can account for the three voltages by superposing three solutions of the above form, suitably modified to account for the different locations of the boundary potentials. Since we want V at the center of a square trough, it no longer matters on what boundary each of the given potentials is, and we can simply write: .. In Fig. ., change the right side so that the potential varies linearly from at the bottom of that side to V at the top. Solve for the potential at the center of the trough: Since the potential reaches ero periodically in y and also is ero at x , we use the form: .6. If X is a function of x and X′′ x X X , assume a solution in the form of an infinite power series and determine numerical values for a to a if a and a : The series solution will be of the form: .. It is known that V XY is a solution of Laplace’s equation, where X is a function of x alone, and Y is a function of y alone. Determine which of the following potential function are also solutions of Laplace’s equation: .. Assume a product solution of Laplace’s equation in cylindrical coordinates, V PF , where V is not a function of , P is a function only of , and F is a function only of . a Obtain the two separated equations if the separation constant is n. Select the sign of n so that the solution of the equation leads to trigonometric functions: Begin with Laplace’s equation in cylindrical coordinates, in which there is no variation: The equation is now grouped into two parts as shown, each a function of only one of the two variables; each is set equal to plus or minus n, as indicated. The equation now becomes Note that n is required to be an integer, since physically, the solution must repeat itself every radians in . If n , then .. Two semiinfinite filaments on the axis lie in the regions ∞ a note typographical error in problem statement and a . Each carries a current I in the a direction. a Calculate H as a function of and at : One way to do this is to use the field from an infinite line and subtract from it that portion of the field that would arise from the current segment at a a, found from the BiotSavart law. Thus .. Two semiinfinite filaments on the axis lie in the regions ∞ a note typographical error in problem statement and a . Each carries a current I in the a direction. a Calculate H as a function of and at : One way to do this is to use the field from an infinite line and subtract from it that portion of the field that would arise from the current segment at a a, found from the BiotSavart law. Thus b What value of a will cause the magnitude of H at , , to be onehalf the value obtained for an infinite filament? We require A filament is formed into a circle of radius a, centered at the origin in the plane . It carries a current I in the a direction. Find H at the origin: We use the BiotSavart law, which in this case becomes: A filament of the same length is shaped into a square in the plane. The sides are parallel to the coordinate axes and a current I flows in the general a direction. Again, find H at the origin: Since the loop is the same length, its perimeter is a, and so each of the four sides is of length a. Using symmetry, we can find the magnetic field at the origin associated with each of the halfsides extending from to a along each coordinate direction and multiply the result by : Taking one of the segments in the y direction, the BiotSavart law becomes The parallel filamentary conductors shown in Fig. . lie in free space. Plot H versus y, y , along the line x , : We need an expression for H in cartesian coordinates. We can start with the known H in cylindrical for an infinite filament along the axis: H I a , which we transform to cartesian to obtain If we now rotate the filament so that it lies along the x axis, with current flowing in positive x, we obtain the field from the above expression by replacing x with y and y with : Now, with two filaments, displaced from the x axis to lie at y , and with the current directions as shown in the figure, we use the previous expression to write At this point we need to be especially careful. Note that we are integrating a vector with an a component around a complete circle, where the vector has no dependence. This sum of all a c [Show More]

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